2
=
•(2 )
To fin d S.D. and Variance : T he second m om ent about origin : at>
I x2.f(x)dx
Mi’ -
6 1 = J x \ (& T ^ ) dx
[ by ( 1 )]
__ L_ [63- a3] (6 - o )
3
( b - a ) (6 2 +, a_2 +ab) “ ( b - a ) ■ ---------------T = •! ( a 2 + 62 + ab).
...(3 )
We know that Variance = jij' - (u i1)2 = |
(a2 + 6^ + ab) -
= g ( a 2 + 6* + ab) - ^ (a2 + 62 + 2ab)
[by (2 )]
4 5 4 | E n o n b e h n g Mathematics-111
- a 2 + b2 - 2 ab 12 _ (b-a)2
[v * >
12 and S.D.
a]
o' = V V a r ia n c e
yfl2.
(b -a ).
5.19 MEAN DEVIATION OF A RECTANGULAR DISTRIBUTION: T he p.d.f. o f rectangular distribution is given by 1
a
b-a
fix)
0,
othew ise
•d )
We know th at m ean deviation ab o u t m ean is 00 M .D . =
}x - m ean \ f ( x ) dx
J
x -
(a + b)
_____ L _ f (fc~ a> I
P u t/ = x -
( a + b)
and lim its are : t => -
x-
[by def.]
1
( b - a ) dx (a + b)
( o + 6) , [ v m ean = — ^— ]
dx
dt - d x
=>
(b-a) ^ (b-a) to
■>K>M .D .
-
(b-a) 1
(b-a)
j
W dt
(b-a)
2
(±±) 2 (b-a)
? J
^
[ v jf| is an e v e n /* .]
S
2
(fr-o) ?
.
I
tdt
= (6 - 0)
tatistics
| 455
[ v |/| = /, if (b - a) > 0 ]
(6 -o )
= __2_
1 r a ]—
( b - a ) ' 2 I Jo 1
(b -a )*
4
(b -a ) • ,
0 -°>
4 ’ Exam ple 5.68 : A variate X has the rectangular or uniform distribution with p .d f. L given by :
1 r
0 < x <100
100 0, otherw ise Com pute (i) P [ X > 60} and (ii) P [20 * X * 4 0 ] m
-
Solution : (i) We have too
P [X> 60] = J f ( x ) dx 60 &
^
100
= I
10 0 *
[v by given/(x)]
100
~ 11 000 W l* J60
= 0.4 (ii) We have, 40
P [20 5 x 5 40] = J f( x ) d x 20
40 = 1 20
=
A
100 *
1 f ]40 10 0 ' XJ20
20 100 =
0.2
A ns.
4 5 6 ( E n g in ee r in g M a th e m a tic s -H !
5.20 GAMA DISTRIBUTION: (i) A c o n tir io u s ra n d o m v a ria b le X is s a id to be a gam a distribution w ith param eter
X > 0, if its p.d.f. is given by : e -V -'
0 <
jc
< oo,
X >0
~ W
/(* > «
0,
othew ise
J e~zxx~'dx. 0 (ii) The p.d.f. a gam a distributin w ith tw o param eters a > 0 and X > 0, is given by Where
[(Ij =
° f i x ) o r f { x , a, \ ) =
0
g^ayX-i •
a > 0, X > 0
Iw
0,
‘
otherw ise
5.21 DISTRIBUTION FUNCTION OF GAMA DISTRIBUTION The p .d .f o f the gam a distribution is 0 < x < c o , A_ > 0
fix)
m 0
; •••(J)othe
We have, distributive function o r C.D.F. o f / (x) is given by
F (x) = P iX < x) = 1 f i x ) dx -00
x( e - '.x x-' F i x ) - J - r ^ r - dx
[by (l)]
1(A)
o
1
[by definition]
j e~*xM dx
W) x
1 H ence d .f
f(X )
F{x) =
/
o 0,
e~* xK~'dx,
0 < x < oo otherw ise.
5.22 MEAN, S.D. AND VARIANCE OF GAMA DISTRIBUTION: T he p.d.f. o f gam a distribution is given by
e~*.xk‘‘ /(* )=
.
m 0,
00 otherw ise
..( l)
S tatistics | 4 5 7
To find mean : We have M ean = fit’ = E (x) <30
= J xf(x)dx —00
7 e 'V '' = \ x 0 RT) I
[by (l)]
dx
00
-T—» J e * x* '- l dx m
o
[v
CO « = J e*. J f'1
[ v |( » + 1) = « [(n)3 ■ W = X. 7o fin d S.D. and variance : T he second m om ent about origin : <50
ji2’ = J x * .f(x ) dx 7
i
-
,
[by (1)3
* [by def. o f G am a function]
[v |(n+l) =«((w)] = X2 + X. We know that V ariance ( n 2) = to ' - (Mi')2 V ariance ( |i 2) = X
K + X - (V = X
4 5 8 | E n g m eem n g M a tv e m a tk s -III
and S.D.
a — -^Variance
This show s that mean and variance are equal fo r the gama distribution.
5.23 BETA DISTRIBUTION OF FIRST KIN D: T h e ra n d o m v a r i a b le X is s a id to b e Beta d istribution o f fir s t k in d , i f its p.d.f. is given by 1 -5 ( a , p ) ■xa~l ^
_
f ( x ) or f i x , a , P) =
0,
0 < x < 1 and a , 0 > 0 otherw ise.
W here B ( a , 0 ) is a B eta function. I R e m a rk :
(1 - x p -' dx
B ( a , P) = f o
fwl¥) |( a + P)
5.24 COMMULATIVE DISTRIBUTIVE FUNCTION OF BETA DISTRIBUTION OF FIRST KIND: The p .d f. o f B eta distribution o f first kind is given by : 1
5 < a rp)
/(* ) =
0 - ^ l.
0,
0 < x < 1, a, p > 0 otherw ise
By definition o f distrubtive function
F i x ) =P{ X<. X) “
/ fix)dx -ao
= I T t k $ xa~ ' v ~ x r '
dx
= 5 ( ^ p ) I r -1 0
*
[by (i)]
Hence d.f. is X
F{ x ) =
j /*-1 (1 - f ) M dt, 1 ;
0 £x < I x £ 1.
N ote : The above cum ulative distributive function f i x ) is called Incomplete Beta Junction, which is giving by equation (2 ).
S ta tistics | 459
5.25 MEAN S.D. AND VARIANCE OF BETA DISTRIBUTION OF FIRST KIND: The p.d.f. o f B eta distribution o f first kind is given by 1 3 < a ,p )
/< * ) =
0 < x < 1; a, P > 0
0,
...( 1 )
otherw ise
To fin d mean : We have M ean - fi\ = E (x) 00
= J x/(x) dx i? ( a ,p )
[by definition]
j x**'-» (1 -JL )M dx 0
^ p j * B
[by ( 1 )]
( a + i, P)
( v J x^~l (1 - x ^ d x - - B ( l , m)]
V B (/, m) =
( a + p ) . a [(a )
[v
[(a ) ( a + p ) |( a + P) a a +P
|(/ + m)
(n + 1 ) - n («)]
- ( 2)
To fin d S.D. and Variance : T he second m om ent ab o u t origin H2
=j
x>f { x ) dx
*(a,p) jJ0 x2. jf“-'
(1 - x ) ^ - ‘ dx
= 5 (^ ~ p ) * B ( a + 2, P)
l( a + P + 2)
[by (1)]
[by def. o f B eta function]
4 6 0 | E n g in eerin g M athem atics -U I
IjEH*---- (g+l)«|(5 R o)
(a + (S + IX a + P ) ^ a + P )
CT^-X, l(
1
)'
a (a + 1)
••(3)
(a + p)(a + p + l) We have Variance (n 2) = M2 - (Mi')2
a (a + l) (a + p)(a + p + l)
a Va+p
[by 2 & 3]
<*• a +1 a a + p j .a + p + 1 a + p « f(a + p) ( a + 1)- a (a + p + l) " a+p [ (a + p) (a + p + l) ap
(a+ p) (a + p + l) S.D. (
and
ap
1
(a + p) V(a + p +1) 5.26 BETA DISTRIBUTION OF SECOND K IN D : The random variable X is said to be Beta distribution o f second kind, if its p .d f. is given by
.
0 -1
1
S M
/(* ) =
'm
S
0
;
0 < i < » a n d a , p >0
;
otherw ise
...(I)
R e m a rk : I f w e put y ~ r -— , then distribution ( I ) becom es, Beta distribution o f first kind.
Exampla 5.69 : Prove that total area (total probability) o f Beta distribution o f second kind is 1. Solution :
The p.d.f. o f B eta distribution o f second kind is given by : 1
fix) =
00
B < a , P ) '( i + x )a+P
otherwise
0,
We have
Total area =
=
[by definition}
J f(x)dx
0}1 B (a, P) (1 + x)a+p
...(1)
dx
[by (I)]
S ta tistics | 4 6 1
a-1 * -
-
j
B ( a ,p ) J ( l + x)a+P
dx
1 1 1-y ~dy [Putting y = -:— - => 1 + x = ~ = > x = —~r~ =>dx = 2 and lim it y~-> 1 (w h e n x = 0) 1+x y y y to y —* 0 (w hen x = 00)] Total area - _ _ L _ B ( a ,p )
J (1
p .:
1 I -d y V j Y ^ 'U 2 J
f fiz z ) 1 I y )
y 5-1
B ( a , P) 0 1 b7
1 (I - > ) a"‘ dy
I
B ( a ,p ) 0,
1
—
B ( a , p)
p ( a , P)
[ v B (/. * ) = J x '- ‘ (1 - or> -1 dx] 0
Hence total area = 1.
5.27 MENA, S.D. AND VARIANCE OF BETA DISTRIBUTION OF SECOND KIND: The p.d.f. o f B eta distribution o f second kind is given by : .'*-1
1
0 < x < co and a , p > 0.
S ( a , P ) '( W . - c f -3
/(* ) =
otherw ise
0 ,
...( 1 )
To fin d mean : We have m ean = ji|' = E (A) 00
=
1 * ( a ,p )
*
1+x
X =
1- y
x a_*
[ * ' ( l + x )a ^
1
[Putting y =
[by definition]
i x f (x) dx
*
dx
dx = - —7 dy and limits o f v -> 1 to 0] v"
4 6 2 | E n g in e e r s M athem atics-I 11
l
? (i - y ) a.(y )a+>
i
B ( “a , pe) ) = ^
a
f
&
0 -y Y * x~' dy
' p j x B (P - 1, a + 1)
W )W )
,.2
[by def. o f B eta function]
](a + p >
l(P —1) - a ( a ) l(a X (P -l) (P -D
■
a
w hen P > I
P -l' Hence
a >*» “ F T
meai
To fin d S.D. and Variance : The second m om ent about origin is given by : 'GO Hz' =
/ x » / ( r ) dx -
_ |
1
^ a -1 [b y (1 )]
5 ( a , p ) x2- ( i + x ) a+p *
£ ( a ,p )
o ( l + * ) a+p
1 l —V _■» [Putting y = — — =>x= —— => a&c = —A a y and lim its o f y A * y y2
1 to 0]
S ta ti s tic s | 4 6 3
’ B lk j)
‘ B (h >
= B ( a,p) xB0 - 2 . a + 2) j f o + P) .. R F 5 j 1( ^ 2 ) [ ( * jR M
l(a + P )
( a + l ) a l a |(p - 2 ) “
|( a ) 0 - 1) 0 - 2 ) |0 - 2 )
, _ <*(<* + !) Mz ( p - l ) 0 - 2 )
m " (3)
We have Variance (jij) = n 2' - ( u i1)2
+ " < p -l)(p - 2 ) g f (g + l) "(p -l)L (p -2 )
f a U -lJ a | (P tfJ
g (a + p -l) ~ ( p - l ) z (p-2) _
and S.D.
[by 2 and 3]
-(4)
a — ^ V a ria n c e
_
1
/a (a + p - 1)~
" 0 -D V
( P “ 2)
5.28 EXPONENTIAL DISTRIBUTION: A random variable X is said to be an exponential distribution w ith param eter X > 0, if its p.df. is given b y :
/«- f Xr": 0;
xS0
otherw ise
...( 1 )
464 |
E nginkfrtng
Mathematks-W
The graph o f exponentai) distribution is
R a m srk :
The total area o f exponential distribution is unity.
S olu tio n :
We have,
1 f{x)dx
[by definition]
=oj Xe-^dx
[by distribution]
total Area =
.-Xx
~x
5.29 DISTRIBUTIVE FUNCTION OF EXPONENTIAL DISTRIBUTION: The p.d.f. o f exponential distribution is given by
Xe-**;
/(* > =
x2Q
0;
otherw ise
...( 1 )
By definition o f distribution function p r C .D .F .: X
F(x)=P(X<,x)= 0
! f(x ) d x *
1 f ( x ) d x + J f( x ) d x 0
—eo
= 0 + J Xe^xxxix
[■j i 1] :J_
[by 0)1
S tatistics I 4 6 5 Hence distributive fu n c tio n :
0, 1-
F (*) =
jc
er*',
<0
jc > 0,
1,
X> 0
JC = ao
Graph o f distribution Function is
5.30 MEAN, S.D. AND VARIANCE OF EXPONENTIAL DISTRIBUTION: The p.df. o f exponential distribution is given by :
m
_
j
le -* * ;
i.
0 ;
xzo,
.
otherw ise
X > 0. ...( 1 )
To find mean : We have,
M ean = ji|* = E
=
( jc)
xf(x)dx
[by defh.]
! x. Xe-^dx
[by ( 1 )]
/
o \00 = X
JC.
-X
oo
-Xx
-S ^ -d x /o
0 + I e -^ d x o
-X
i~Xx -X
- X ( ° - !>= X Hence
1 jii = m ean = ^
••(2 )
4 6 6 j E n g in eerin g M athematics -III
To find. S.D. and Variance : T he second m om ent about origin
J x\f(x)dx
-
= J x2.Xe~** dx
[by (i)I
- X J x2.e~k*dx
= X
x 2. e
0”
-x L
; 0 + 2 J xe~u dx 0
= 2
00
I 0
„
' Hence
\
X* ( 0 ~ ° ~
I *
■ _2 _
/0 _
X2
2 “ X2
...(3)
We have V ariance fi2' - ^ 2^ - (Mi')2
2
-* -« r and
S.D. : a - 1/ V a r ia n c e ~ T
[by 2 & 3]
S tatistics | 4 6 7
5.31 MEAN DEVIATION ABOUT MEAN OF EXPOBEBTIAL DISTRIBUTION: The p.df. o f an exponential distribution is given by /< * ) =
Xe-**; 0
x'i.O and X > 0
;
otherw ise
- .( I )
We know that m ean deviation about m ean is given by M .D . =
J
l x - m e a n |/ ( * ) d r
[by ( 1 ) and mean ~ jj-]
= J |Xjc —1[ er^dx 0
Puting Xx = t => Xdx - dt and limit t -> 0 to «>, w e have
go M .D . =
Since
I '- H
f i \ t - \ \ e-
=
f1 i
X
°o
] (1 - i) e~ld t + J (* - 1) e ' d t o
l
= X [ - ( 0 - D + ( O o - ( 0 - 0) - ( * r O r ] X [1 + (
*r , Xr ‘
[by
2)
468
| E n g in e e r in g
Mathematics-411
Example 5.70: A random variable X has an exponential distribution w ith p.
.
I[ 0,
i f x SO . Com pute th e probability th a tX is not less than 3. A lso fin d , m ean and standard deviation and prove th a t coefficient o f variation is unity.
Solution : (i) Com pute P (X K 3) : We have
P (X * 3) = P (X > 3) 00 = / f(x)dx
3
'
00
= J 2 e-2* dx
3
= - [0 - e~*] = (ii) To fin d m ean : We have,
m ean = fi|' = E (x)
S tatistics | 4 6 9
—
5 (0 -1 )
_ I 2
(iii) To fin d S.D . :
We have,
1
H:’ =
-C O
t
= J x2. 2e-2* cfr
= 2 J x'er^dx 0
3- e
-2x
-2 x V
- ! 2x S ~ d x
"2Jo o
=2
2 0 - 2 J x e 2xd x ( - 2) o
CO
= 2 / x
~2
4 7 0 | E n g in e e rin g M a th e m a tic s-!!!
- j er^dx o
e - 2* )*
- 2 Jo = - 1 (0 - 0 We know that, S . D . : a - ^jj-2 ' - ( m ') 2
■
f W
=
/H I V2 4
- UH =I ~ V
and
4 ‘ 2
Variance = ^
A ns.
To fin d Coefficient o f Variation :
We have that, coefficient o f variation =
„_,
1/2 = 172=1.
^ved.
Example 5.71 : The Incom e ta x X , o f a m an has an exponential distribution w ith p .d f. is given by
f(x )
0
;
x <0
I f incom e ta x is levied a t th e rate o f S% , w hat is the probability th a t h is Incom e exceed Rs. 10,000 ?
Solution : If income exceed Rs.
10,000, then income tax will be : Income tax = 10,000 * = 500.
S tatistics | 4 7 1 H ence tax o f a m an will exceed Rs. 500. required p ro b a b ility :
P (X > 500) =
\ f(x)dx
500 00
=
7 e~*IAdx
J
[by given p.d.f]
600 4 aO
i [ l -*n y I 4 1 -1 /4 500 = _ (0 - e-[” ) = e~]2S
A ns.
E x a m p le i5.72 : Suppose Ike life o f mobile batteries is exponentially distributed with parameter
X —.001 days. What is the probability that a battery will last more then 1200 days ? S olution :
The p.d.f. o f exponential d is trib u tio n :
fix)-
Xe~^ \
x >0
0
otherw ise
;
...( 1 )
w here X, = 0.001 days.
P (battery will last m ore than 1200 days) = P ( X > 1200)
=
I f ( x ) dx 1200
J
hr^dx
1200
( * rV
[using ( 1 )]
J\2< '1200
=_ (0 _ {j-XxIMO)
: ^-0.001*1200 e~12 « 0.301
[X = 0.001] A ns.
4 7 2 | Engineering Mathematjcs-JH
5.32 TEST OF SIGNIFICANCE The statistical co nstants for the given population such as m ean (p), standard deviation ( a ) etc. are called the parameter's. Sim ilarly, the statistical constants for the sam ple draw n from the given population such as m ean ( ic ), standard deviation (j) etc. are called th e 'statistics'. In general, the population param eters are not know and th eir estim ates given by the corresponding sam ple statistic are used. It is necessary to test th e statistic to see w hether their difference is significant or not. Such tests are called th e 'tests o f significance'. It can be used to com pare the characteristics o f tw o samples o f the sam e type. For applying th e test o f significance w e first setup a hypothesis.
5.33 NULL AND ALTERNATIVE HYPOTHESIS A hypothesis w hich is tested for possible rejection under the assum ption that is true is called null
hypothesis, and is denoted by HoIn other w ords, a null hypothesis Ho that there is no variation exist betw een variables or set of given observations.
For example, T he null hypothesis H0 m ight be that there is no relationship betw een tw o measured phenom ena. The hypothesis w hich is accepted, w hen the null hypothesis has been rejected is called the altenative hypothesis, and denoted by H) o r HA For example-, in a clinical trial o f a new drug. Then, w e have Null hypothesis Ho : there is no difference betw een the tw o drugs on average i.e; the new drug is no better on average, than th e current drug. A lternative hypothesis H\ : the tw o drugs have different effects on average, i.e; the new drug is better on average, than the current drug. The purpose o f hypothesis testing is not to question the com puted value o f the sam ple statistic but to m ake a judgm ent about th e difference betw een that sam ple statistic and a hypothesized population param eter.
534 LEVEL OF SIGNIFICANCE The probability level below w hich are reject the hypothesis is know n as the level o f significance. T he level o f significance usually em ployed in testing o f hypothesis are, 5% and 1%. W h a t, if we test a hypothesis at the 5% level o f significance? T his m eans that w e will reject the null hypothesis, if the difference betw een th e sam ple statistic and population param eter is so large o r a larger difference w ould occur, on th e average, only fine o r few er tim es in every 100 sam ples when the population param eter is correct. A ssum ing the hypothesis is correct, then the significance level indicates the percentage o f sam ple m ean that is outside certain limits. In this section we shall study the follow ing four tests o f significance only, viz : (i) x 2 - test, (ii) t - test, (iii) F - test and (iv) Z - test.
5.35 CHI-SQUARE TEST OR x2- STATISTIC The C hi-square test is a very pow erful test fo r testing the significance o f the discrepancy between the actual (or observed) frequencies and the theoretical (o r expected) frequencies in a sample, ft w as given by prof. Karl P earson in 1900 and is know n a s " Chi-square test o f goodness offit". The C hi-square is a letter o f th e G reek alphabet w hich is w ritten as %2.
S ta ti s tic s | 473
Let f 0 and f be a se t o f observed an d expected frequencies o f a class interval o r cells. Then Chisquare distribution is defined as ifo-fe)2
- ( 1)
X2 W here 'Lf„-'Lft = N = total frequencies
Another Form o f %z distribution :
k /w .r l Since
X2 = 2
I
I
[/„ *
I - I
'•
{£-% ♦/.[ f 2 /o fe 4
f 2 *0 -2 N +N
or
<
[v i/0 = i / , = ao
...(2 )
-AT
5.36 PROPERTIES O F x2 DISTRIBUTION : (i)
The values o f %2 are alw ays positive, i.e; x 3-c u rv e is alw ays positively skewed.
(ii)
T he low est valu e o f %2 is zero and the highest value is infinity.
(iii)
I f degree o f freedom v = 1 , then x ^ u r v e will be y = y 0e~*
, w hich is the exponentail
distribution as show n the follow ing figure
(iv)
I f v > 1, then x *-curve is tangential to X -ax is at the origin and is positively skewed.
474
( E n g in e e r in g M a th e m a tic s - ! ! ]
(v )
T he probability P th at the value o f y} from a random sam ple will exceed xi >s gives by
P = J ydx. *0
537 CONDITIONS FOR THE VALIDITY OF x*-TEST: (i)
T he m em bers o f sam ple should be independent.
(ii)
C onstrains o n th e cell frequencies should be linear.
(iii)
Total frequencies N should be reasonably large say, greater than 50.
(iv)
As x ^ -te st is based w holly on sam ple data, no assum ption is m ade about the population values (or param eters).
(v )
T he expected frequencies should not be less than 5, If it is less than 5, than adjacent it.
(vi)
x 2-test
w holly dependent o n degress o f freedom (do/.) v.
538 USES OR APPLICATION OF x*-TEST: (i)
(ii)
Test o f goodness o f f i t : C h i-s q u a re s ta tis tic m ay b e u se d to d e te rm in e w h eth er tw o curves are fitted good o r not. Test o f independent o f attributes in a contingency table test for a specified standard deviation.
(iii)
Test fo r a specified standard deviation i.e ; it m ay b e u se d to te s t o f p o p u latio n variance.
(iv)
C om pare a nu m b er o f frequency distributions.
Remarks on the x*-Test : 1. C ells : I f given data are arranged in the com partm ents, w hich is called cells or class interval and frequency o f a cell is called cell frequency. 2. Contingency Table : Let a group having N persons is divided into tw o attributes A and B. A having m classes say, A lt A 2........A , ......... and B having n classes say, B u B2, Bi t ......... B„ and
Am
/oij are observed frequencies o f celles belonging to classes A, and Bj.
A \B
B,
B 2 .... ......
A|
f\ l
f \ 2 •— ...... A j .........
/.
a2
fix
h i ...... ......f u ......... ....... f u
fi
A
fix
fa
A*
fm\
fm2 ......
Total
Ft
Fi
Bj ....... ......
...... ...... f j
..... .....
B„
.......... ....... U
Total i
f. fm
Fj
.......
....
Fn
N
S ta ti s tic s | 4 7 5
3. Calculation of Expected Frequencies : C onsider the 2 * 2 contingency table :
A \B
B,
b2
A, a2
/.
h
h
U
N, n2
Total
Hi
Na
iV (G rant Total)
Total ^
r
The expected (o r theoretical) frequency o f cell 11 is given by
t\\ f e\
N
Expected frequency o f cell 12 is : ft\2
-------- J T
Expected frequency o f cell 21 is n
Mi
3*n 2 N
Expected frequency o f cell 22 is :
f
U22
- ^ 4 * ^ 2
N
4. D egree o f fre e d o m (d o f) : T he num ber o f independent variates is called the num ber o f degree o f freedom (dof) and denoted by v. (i)
In Binomial d istrib u tio n : dof v = n - 1, (w here n is total sam pling, for exam ple x : 0, 1, 2, 3 , 4 then n = 5 - 1 = 4 )
(ii)
In poisson distribution : dof v - n - 2
(iii)
In N orm al distribution : dof v = n -
(iv)
3.
For m * n contingency table. d o f v = (m - 1 ) ( n - IX w here m = num ber o f row s, an d
n - num ber o f colum ns.
W o rk in g R u le o f x 1- T est : Step 1. C onsider the Null Hyp*,
s is H 0 : N o association exists betw een the attributes.
Alternative Hypothsis H \ : A n association exists betw een the attributes. S tep 2. C alculate the ex pected (o r theoretic) frequency / „ corresponding to each cell.
476 | E n g in e e rin g M a th e m a tic s-IH Step 3. Calculate x 2-statistic by the formula.
and calculate the degree o f freedom v. S tep 4. See the value o f x 2 from the table, i.e; value o f d o f v, as calculated in step 3 .
at u.% level o f significance and for the
N ote : If no value o f a % is m entioned, then take 5% i.e; a ~ .05 S tep 5. D ecision : (i)
If calculated value o f x~ < tabulated value o f x i >• Then accept the null hypothesis H0.
(ii) If calculated value o f x 2 ^ tabulated value o f xi,* T hen rejected null hypothesis llQi.e; accept the alternative hypothesis H\.
Example 5.73 : From the table given below, whether the colour o f so n ’s eyes is associated with that of father's eyes ? Given that the value o f x 2fo r / d o f at 5% level o f significance is 3.841. Eye Colour is so n ’s Not Light
Light
Eye colour
Mot Light
230
148
in father
Light
151
471
Solution : S tep 1 : N ull H yothesis H0 : T he colour o f the son’s eyes is not associated with the colour o f father's eyes. S tep 2 : C alculation o f Expected Frequencies f < :
Light
Total
Not Light
fo\\ = 230
f , n = 148
378
Light
foi\ - 151
622
Total
381
619
N « lOUO
iv
N ot Light
i!
G iven the O bserved freq u en cies/,// are :
The Expected frequencies will be : N ot Light
N ot Light Light
Light
A"
378 x 381 1(X)0
...
/<21 '
381x622 _ 10 0 " 237
.
}' n
378x619 10 0 0 619x622 10 0 0
_
S ta tis tic s | 4 7 7
S tep 3 : C alculation ol ' x : -stutisiu- • We have or
^ (fo ■ fc ?
x- ■=
2 = < 2 3 0 -1 4 4 )*
( 1 4 8 - 2 3 4 ) 2 (151 - 2 3 7 ) 2 O O A 144 234 2O 3l3 7’7 - 51.36 + 31.61 + 31.21 + 19.21 = 133.29. 1 A A
(471 - 3 8 5 ) 2 g g g
and degree o f freedom V
- (w
- (2
I)
I)
1)(2
I)
[ 2 ^ 2 c o n tin g e n c y ta b le ]
-- I. Step 4 : The calculate value o f X2 at 5% levels o f significance and for 1 d o f is 3.841 i.e.. Xno5.i “ 3 84 I S te» S : Decision : C learly calculate value o f x ’ - 133.39 ^ tabulated value o f Xnovi - 3 84 ro the Null hypothsis is rejected i.e. there is an association betw een thfe colours o f eyes o f son's and colours o f eyes o f father's. A ns.
Example 5.74 : In a sample survey o f public opinion, uswers to the questions fi) Do you drink? (ii) Are you in favour o f local option on sale o f liquor ? are tabulated below : Question Yes
No
Total
Yes
56
31
87
No
18
6
24
Total
74
37
UI
Can you infer whether or not the local option on the sale o f liquor is dependent on individual drink ? (Given that the value o f z 2fo r d o f at 5% level o f significance is 3.841.) S olution :
Step I : Null Hypothesis : The option on the sale o f liquor is not dependent {or not associated) with the individual drinking. S tep 2 : Calculation of expected or theoretical frequencies f e The expected frequenie.s. corresponding to the observed frequencies are calculate as follows Expected frequencx o f cell 11 is 87x74
4 7 8 | E ngin eerin g M athematjcs-111
E xpected frequency o f cell 12 is
f< " = - t i t
5=29
E xpected frequency o f cell 21 is _ 74x24 ~ 111 ~ E xpected frequency o f cel! 2 2 is
x
U i~ S tep 3 : Calculation o f
We have
24x37 m
- 8.
statistic :
x: “ ^
! ^ ~ >e
-(1
ifo-fef
fa
f.
56
58
4/58
31
29
4/29
18
16
4/6
6
8
4/8
U
z !. m
Total H ence
, o _937
Value o f j } = 0.957
A lso degree o f freedom
(dof) v - ( « - 1) (n - 1) = (2 - 1 ) ( 2 - 1 ) = 1.
S tep 4 : T h e tabulated value o f %2 at 5% level o f significace and fo r 1 d o f is 3.841 is 3.841 X0.05,1 = 3.841, Step 5 : Decision: C learly calculated value o f x2 = 0.957 < tabulated value o f x2o.os.i =3.841. => die N ull hypothesis is accepted. => S ale o f liq u er is n o t dependent o r not associated w ith the individual drinking.
A ds.
S ta ti s tic s 14 7 9
Example 5 .7 5 : SOstudents selected at randomfrom 500students enrolled in a computer crash programme were classified according to age and grade points giving the following d a ta : Age (in years)
Grade Points
21-30
above 30
Total
up to 5.0 5.1 to 7.5 7.6 to 10.0
3 8 4
5 7 8
2 5 8
10 20 20
Total
15
20
15
* li *
20 and under
Test at 5% level o f significance the hypothesis that age and grade point are Independent (Given that Xaos,i = 5.99J). S tep 1 : N ull H ypothesis H0 : T he age and grade point are independent. S tep 2 : Calculation o f expected frequencies f t : T he expected frequencies are as follow s : G r a d e P o in ts
20 a n d u n d e r
2 0 x 10 50
5.1 to 7.5
15x20 , 50 “ 6
2 0 x 20 50
7.6 to 10.0
15x20 50
2 0 x 20 50
upto 5
a b o v e 30
2 1 -3 0
o * w 11 UJ , 6
. 4
15x10 50
8
15x20 , 50 =6
'
15x20 50
. 3
.
From the ab ove table, w e see that som e values o f expected frequencies are less than 5, so that to am algam ate these cells to its neighbours (i.e; m erge 1 & U rows). T hen new obseved an d expected frequenies table a re : E x p ected fre q u e n c ie s ta b le (ft) : 20 a n d u n d e r
2 1 -3 0
above 30
I b ta l
upto 7.5 7.6 to 10.
9 6
12 8
9 6
30 20
T otal
15
20
15
Z II g
Solution :
20 & u n d e r
2 1 -3 0
ab o v e 30
Total
upto 7.5
11
12
7
30
7.6 to 10.0
4
S
8
20
T otal
15
20
15
N = 50
G r a d e P o in ts
O b se rv e d fre q u e n c ie s ta b le {f0} : G r a d e P o in ts
4 8 0 | E n g in e e rin g M athematics-111
§ t s n l : Calculation o f x2-statistic : We have,
x2 ~
nr
v» X
9
i ( 1 2 - 1 2 )2 12
( 7 - 9 )2 9
( 4 - 6 )2 6
( 8 - 8)2 , ( 8 - 6)2 8 8
= 2 .222 . A lso degree o f freedom . (dof) v = (m - 1 ) (n - !) = (2 - 1) (3 - I) i.e;
[ v A s new table]
d o.f. • v = 2
S tep 4 : T h e tabulated value o f x 2 at 5% level o f significance and for d o f v = 2 is 5.991 i.e.. X&.m 2 = 5.991. S tg P -5 : Decision : C le a rly c a lc u la te d v alu e o f x 2 = 2 .2 2 2 < ta b u la te d v alue of XSoj.2 = 5.991 => Null hypothesis ts accepted. => the grade points and age are independent o f each other.
Ans.
Exam ple 5.76 : In experiments on pea-breading, mendal obtained the following frequencies o f seeds:
R ound and
W rinkled and
R ound and
W rinkled and
yellow
Yollow
Green
Green
315
101
108
32
Total
556
Theory prtodlcts th a t the freq u en cies should be in proportion 9:3:3:1. E xam ine the correspondance betw een theory and experim ent G iven th a t th e value o f x 1fo r 3 d o f at 5% level o f significance is 7.81S. S olution : S tep t : Null Hypothesis H0 : there is a correspondence betw een theory and experiment. : Calculation o f Expected frequencies ( fj : G iven frequencies in proportion 9 : 3 : 3 : 1 Total sum o f proportion = 9 + 3 + 3 + 1 = 16, therefore (i)
9 E xpected frequency o f Round and Yellow seed is = j g * 556 » 313
(ii)
g E xpected frequency o f W rinkled and Yellow seed is = j g * 556 * 104
3 (iii) E xpected frequency o f R ound and G reen seed is - j g * 556 ~ 104 (iv) E xpected frequency o f W rinkled and G reen seed is =
* 556 * 35.
S ta tis tic s | 481
Step 3 '.Calculation o f %2-statistic :
H ere
f 0 = observed or actual frequencies f e = expected o r theoretical frequencies
r , _ (3 1 5 -3 1 3 )2 (1 0 1 -1 0 4 )2 (1 0 8 -1 0 4 )2 (3 2 -3 5 )2 X 313 104 104 35 ^ 4 9 , 16 , 9 3 1 3 104 104 3 5 c A lso th e degree o f freedom v = n - l = 4 ~ 1 = 3 . S tep 4 : T he tabulated value o f x 2 W 5% level o f significance and for d o f v = 3 is 7 .8 15 X005.3 ~ 7.815 S tep 5 : C alculated value o f %2 = 0.5 < tabulated value o f x 5.os,3 = 7.815 => N ull hypothesis is a accepted. there is very high degree o f correspondence betw een theory and experim ent. A ns. Example 5.77 : A servey o f 320fam ilies with S children each revealed the follow ing distribution :
No. o f boys :
5
4
3
2
1
No. o f g ir ls :
0
I
14
56
3 88
4
No. o f fam ilies :
2 no
0 5
40 12 Is this result consistent with the hypothesis that the male and fem ale births are equally probable ? (Given zks, *-s * 10.07). Solution :
Step 1 : Null Hypothesis Hq : M ale and
fem ale births are equally propable.
S t e a l : Calculation o f Expected frequency f t : From th e Binom ial distribution, th e expectedfrequency:
f (r) = N. "C, q ^'p ”',
r = 0 , 1, 2, 3, 4, 5.
H ere N = 320, p = q = ^ (equal probability) and n = 5. For r = 0, then (1) becom es : / , ( 0 ) = 3 2 0 * 5 C 0 ( ! ) 5 ( ! ) ° “ 10 F or r = 1, then / , ( ! ) = 320 * ^
( I ) '( I /
=50
F or r = 2, then
f t (2 ) = 320 x JC2 ( i )
(i)
= 100
f (3 ) - 320 x SC 3
(!)
=100
For r = 3, then
4 8 2 1ENQMemiG M/utcm/otcs-III For r = 4, then / , (4) = 320 x !C4 ( i ) 4 ( i ) ' = 50 For r ~ 5, then / . (5 ) = 320 x SC5 ( | ) ° ( | ) 5 = 10.
Hence expected frequencies for r = 0, 1,2, 3 ,4 , 5 or 5 ,4 , 3 ,2 , 1,0 a re : / , - 10, 50. 100, 100, 50, 10 : Step 3 : C alculation o f z 2-sta tistic : we have
= Z fe
/.
/,
14
10
56
50 100
110 88 40 12
< f* - f.W
< /;-/.)* '
1.60 1.72
100 50
36 100 144 100
1.44 2.00
10
4
0.40
1.00
Total
%2 = Z
f(/o -/« )* ! ’ = 7.16 fe J
Also degree o f freedom for BD . v " 6 - 1 = 5. S eto4 : The tabulated value o f %2 at 5% level of significance and for dof v - 5 is 11.07 i.e, $ 05,5 ■ it .07 StenS : D ecision: Clearly calculated value of %2 * 7.16 < tabulated value of xS.05,5 * 11.07 => Null hypothesis is accepted => the male and female births are equally probable.
Ans.
Example 5.78: In 60 throw s o f a dice,fa c e one turned up 6 tim es, fa c e tw o or three, 18 tim es, face fo u r or fiv e 24 tim es and fa c e six, 12 tim es, Test a t 10% level o f significance, i f the dice is honest, it being g ives th a t P ( z * > 6*25) ~ 0,1 fo r 3 degrees o f freedom . Solution : Step 1 : N ull H ypothesis Hq\ The dice is honest Step 2 : C alculation o f expectedfrequenies f , : Given that the observed frequency are as follows : Face o f dice x or (r) : 1 %or 3 observed frequency f a
:
6
18
4 or 5
6
24
* 12
(N = 60)
S ta ti s tic s J 4 8 3
Under the assumption of Ho that the dice is honest, so that expected frequency for each face is /.W
“ N x g - 6 0 x i = 10 .
[Because prob. of turning up any one of the numbers J, 2,3,4, 5,6 is g]. i.e. x or r : 1 2 or 3 4 or 5 6 expected frequency f t 10 20 20 10 Step 3 : C alculation o f £ s ta tis tic : Wehave
, - E
— s-
Tkble for x2-statistic : Face of dice x or r
/.
fe
10
2 or 3 4 or 5 6
6 * 18 24 12
Total
tf-6 0
I
u 16 0.2 0.8 0.4
20 20 12 JV*60
x2= £ '
\ i f o ~ f e )2 fe r
3-0
Thus,
Calculated value of %2 * 3.0. Also degrees of freedom v = 4 - 1 =3. Step 4 : Since given that tabulated value of v3at 10% level of significance and for 3 dof is 6.25 i.e. xS.i,v»3 “ 6.25. : D ecisio n : Clearly calculated value of x2 “ 3.0 < tabulated value of XS.i.3 - 6.25. => Null Hypothesis is accepted => The die is honest Ans. am ple 5.79: C alculate th e expected frequencies fo r the fo llo w in g data presum ing the two attributes C ondition o f hom e
C ondition o f ch ild
G ear
D irty
dea n
70
50
F airly clean
80
20
D irty
35
45
UsecM*quaretestat5%lmldf9ignyteancelosm6 whethcrthetwoattributesareindependent (table value o f£ a t 5%for 2 dcfit 5,991, endfor 3 dofis 7,915 and fo r 4 d o f Is 9.488).
4 8 4 | ENQBoramo M m h em m k s -111
Solution i
Step 1 : N ull H ypothesis H q : No association between the attributes. Step 2 : C alculation o f expected frequency ( fj o f each c e ll :
Clean
Dirty
f o : 70
f o ' 50
120x185 300 /0 :8 0
Fairly Clean *e
Dirty
Clean
185x100 300 fo : 35
,
1
f t =
, Total
185
„ = 4 6
115x100 300 fo * 45
1
/ • - 1SIZ om = 49*33
115x120 ~ m " / . : 20
Total
115x80 ____ 300 = 30-67 115
120
100
80 N = 300
Step 3 : C alculation o f x 2 -sta tistic :
_ ( 7 0 - 7 4 ) 2 . (50 - 46)2 . ( 8 0 -6 L 6 7 )2 . (2 0 -3 & 3 3)2 X2 “ - 7 4 - ’+ — 4 6 - + ~ L67— + “ W ‘ t ( 3 5 - 4 9 3 3 )2 (20 - 38.67)2 4933 + 30.67 •» 0.2162 + 0.3478 + 5.4482 + 8.7657 + 4.1627 + 6.6954 = 25.636. Also degree o f freedom (d.o.f.): v = (m „ I) (n - 1) = (3 - 1) (2 - 1) = 2. Step 4 : Since the tabulated value o f %2 ** 5% level o f significance and for dof v = 2 is 5.991, i.e; xfl.ov * 5.991. Step 5 : D ecision : Clearly, calculated value o f x2 * 25.636 tabulated value of X&.05, 2 = 5.991 => the Nufl Hypothesis is rejected. => there exists and association between the attributes. Ans. Example 5.80 : The fo llo w in g fig u re s show the distribution o f digits in num ber chosen a t random fro m a telephone d irecto ry: D ig its: 0 1 2 3 4 5 6 7 8 9 F reuency: 1026 1107 997 966 1075 933 1107 972 964 853. Test a t 5% level w hether th e digit* r m y b f Ufken to occur equally freq u en tly in the directory. (T he ta b le w in e o f fo r 9 degree o f freedom m 16.919)
S tatistics I 4 8 5
Step 1 : Null Hypothesis H o : The digits occur equally frequently in the directory. Step 2 : C alculation o f expected frequency ( fj : V Given that total observed frequencies N * l f Q“ 10,000. Expected frequency for each digits 0, V, 2 , ..... 9; f.{ r)~ N * p { r)
10,000 x ~
I* p (r) = to ]
for r = 0, 1, 2,..., 10.
* 1000;
Step 3 : C alculation o f -sta tistic : f2 *-
We have,
I
xorr
f.
/. .
0 1 2 3 4 5 6 7 8 9
1026 1107 997 966 1075 933 1107 972 964 853
1000 1000 1000 1000 1000 1000 10Q0 1000 1000 J m .
N = 10,000
N = 10,000
(fo -fe )* fe
0.676 11.449 0.009 1.156 5.625 4.489 11.449 0.784 1.296 21.609
.
k / w , ) 2! ■jk—58.542 le
Also degree o f free& ^ v ^ 10 - f 1* 9. Steiy 4 : Tfife tabulated vafue o f x* a?5% level of significance and for dof v = 9 is 16.919, i-e;$o5,9 16.919: Step 5 : D ecision : Clearly, calculated value of x2 * 58.542 * tabulated value of xios.9 = 16.919. => the Null Hypothesis is rejected. Ans.
=> the digits do not occur equally frequently. Exam ple 5.81: A d ice is tossed 120 tim e* w ith th e fo llo w in g re su lts: N um ber turned u p : F reuency:
1
2
$
4
5
30
25
18
10
22
Test th e hypothesis th a t th e dice is unbiased (fits , s m 11*07)
6 15
Total
120
4 8 6 { E fin n rm a
-V
ig
M athem atics- i n
Solution :■ Step I ; Nutt Hypothesis Hq : The dice is unbiased oat. Step 2 : Calculation o f expectedfrequency : Here N 31120. On the basis of Hq, the dice is unbiased* so that the expected frequency / , ( r ) - N p ( / 0 - 120 » | - 20; for r - 1 ,2 ,....6 i.e; expected frequencies for each turned up r * 1,2 ,3 ,4 ,5 ,6 is 20. Step 3 : Calculation o f ^sta tistic : We have
... ( 1)
Table for x2-«tatistic Cf . - f #
(f o - f e )■ fe
20 20 20 . 20 20 20
100 25 4 100 4 25
5.00 1.25 0.20 5.00 0.20 1.25
tf»120
—
X2 - 12.90
x o rr
/.
f.
1 2 3 4 5 6
30 25 18 10 22 15
Total
AT-120
Also degree of freedom v = 6 - 1 =*5. Step 4 : The tabulated value ofx2 at 5%level of significance and for dof v=5is 11.07, i.e; Xtas = 11.07. Step 5 : Decision: Clearly, calculatedvalue of x2* 12.90 K tabulated value of xS.os.j * 11 07. =>the Null Hypothesis is rejected. => the dice is a biased one. Ans. Example 5.82 : Front the adult male population c f seven large cities random sample giving 2 x 7 condnguency table cf married andunmarried mem ra given below were taken. Can it be said that there is a significant variation among the cities In the tendency of men to marry T City A B E C D F G Total Married 133 164 155 106 123 153 146 980 Unmarried 36 57 40 37 55 33 294 36 Total 169 195 221 143 208 156 182 1274 (At (2 —1) (7 —1) d.f. Hike 12.6) Solution : Step 1 : Null Hypothesis H0 : There is no significant variation among the cities in the tendency o f men to many.
S tatistics 1 4 8 7
Step 2 : Calculation of Expected frequency: On the basis of Null hypothesis the expected frequencies are: 980 Expected number of married people in City A * J2 7 4 x *69 = 130 980 Expected number of married people in City B * J2 7 4 x 221 =
170
980 Expected number of married people in City C * J2 7 4 * ^ = 15® Expected number of married people in City D ~
980
x
“
=
980 Expected number of married people in City E ~ J2 7 4 x 208 - 160 Expected number of married people in City/*=
980
1274
x 156 ~ 120
980 Expected number of married people in City G m J2 7 4 x 182 = *40 Similarity the Expected frequency are 39, 51,45,33,48,36 and 42. T a b k for expected freqieacy f 4 Married 130 170 150 110 160 Unmmarried 39 51 45 33 48 Step 3 : Calculation of -statistic :
120
36
We have, /. 133 164 155 106 153 123 146 36 57 40 37 55 33 36 N= 1274.
-
-
A
V .-f#
130 -THT 150
9 36 25 16 49 9 36 9 36 25 16 49 9 36
110
160 120
140 39 51 45 33 48 36 42 N** 1274
fe 0.069 0.212
0.167 0.145 0.306 0.075 0.257 0.231 0.706 0.556 0.485 1.021
0.250 0.857 X2 = 5.337
140 42
4 8 8 | E n g m ee m n g M a ih em * iic s -H1
m m i • Th* tabulat&d value of x2 a* 5% level of significance and for v = (2 - 1) (7 -1) = 6 dof is 12.6 i.e., xfl.os, 6 * 12.6 . Step 5 i Decision : Clearly calculate value of x 2 * 5.337 < tabulated value of X2 * 12.6 => the null hypothesis is accepted. => these is no significant variation among the cities in the tendency of men to marry. Example 5.83: For 2 * 2 contingency table
(o + b + c + d ) ( a d - be ) 8 z - ( a + b )fr+ ,rf(b+d) ia + cy
prove that,
Solution : Given the observed frequency (fa) table :
Total
a c
b d
*+-c
b+a
Total a+b c +d
i
|
N=a + b + c +d
Table for expected frequencyf e :
(a+6)(6+d)
(o + 6 )(a + c) a+b+c+d (c+
o + 6 + c+ d (c+d)(]b+d) o + 6 +c+d a w . ) 2!
Wehave,
E
r l
(o+ 6 )(« + c)i 2 a+b+c+d J (q+ 6)(q+ c) q+6+c+d
[As four terms]
- V ' [q2 + ofr+qe-fad-q 2 - a c - a b ^
mV
(q + 6) ( q + c ) (q + 6 + c + d)
(a d -b e)2______ ( q + 6 ) (q + c) (in- 6 + c +d)
[As four terms]
Now open the summation for four terms, we have r2 -
(qd-fcc) L_r____ i b)(c * c) (q + 6 +c+ d) L(q+W
J*
(q + b)(b+ d)
(c+d)(a + c)
(c+d)(b+d)
S tatistics 1 4 8 9
{a d -b e)2 ff b+d+a+c \ .f b+d + a + c 11 [ l( a + 6 Ko+ c)(b+d) r U' l(c+ c + ddXXoo+ + ccXb+d)!} (a + b + c + d ) Ll(a+&Xa+c)(t>+d)J )(6 -(a d -b e y
[(a + b )(o + c)(b + d ) + fc + d X « + c )(6 + d )] _______ c + d
+ a + b _______ 1
(a+b)(c+d)(b* d)(a +c)J
( a + b + c + d ) (a d -b c)2 (a + b)(c+d)(b+d)(a+ c)
Proved
Example 5*84: Prove thatfor a 2 * n contingency table:
x ^ fi+ ft wherefi, f 2 are twofrequencies in a subgroup and Nb rows. Solution : Table for observed freqaeacy / , : ----- 1-----------
are the marginal sums of two
Marginal sum (Total)
-
h
Marginal sum (Total)
N,
»•••
f\
N* N, + N2
fi+ fi
The expected frequency corresponding to f , : ( / . + / 2) * i n, +n2 ( / l + / 2) * 2 n,+ n2 ' We have
%2 = £
^ y -^ -
Taking value of chi-square for the first column are :
{ f i+ fi) N \ Nt +N2
C/i + f i ) ^ 2 n x+ n 2
[ v by (I)]
4 9 0 1 En g m k r m g M ath em atics-!!!
iVi(/i+/2) W + N 2) W2(/, +f 2XN1+N t ) =
1v (a " 6)2 = <*'
(A + /*)W + iV 2)
,XM M AW ,
n ,n
2 \& __-^-T 1 2 LM N t l (fl+ fl)
Hence the value of chi-square for 2 * n table is
/T >
Proved.
/ l + /2
Example 5.85 : /» jfte accounting department o f a bank 100 accounts are selected at random and examinedfor errors. Thefollowing results have been obtained. No. o f Errors 0 1 2 3 4 5 6 No. rtfAccounts 36 40 19 2 0 2 1 DoestUs information verify that the errorsare distributedaccerdlng to Poisssonprobability tow 't (tbs, s m 11.07). Sobttkm ; Stop 1 : Null Hypothesis Ho : The errors are distributed according to Poisson probability law. S teal : Calculation of expectedfrequency (£): We know that the expected frequency of poisson distribution : e~m mr fA r)= N - — where
’
0, 1, 2,3,4, 5, 6 .
N = y ~ 100 Zfr m = mean = - Qx 36-flx40-h2xl9 + 3x2 + 4xQ-hSx2-f6xl m~ 36 + 40 + 19 + 2 + 0 + 2 + 1 _ 100
100 " *•
Expected frequency: , 1°
100x0367
S ta tistics | 491
37
For /• = 1 :
^ (1) = 100 x r i. ffi!. *
For r = 2:
/, (2) = 100 *
Forr=3:
f. (3) - 100 x e->. S i-
Forr~ 4 :
/ , (4) » 100 x e~K
For r ~ 5 :
o )5 f t (5) * 100 * e~l. "jj" = 0.306 » 0
For r -
f, (6 ) * 100 x
:
6
^ « 18 [2
(i)4 14
0)‘
«6
«2
* 0.051 « 0
Step 3 ; Calculation of £ statistic : We have
X2 =
-< l) /* Since same expected frequencies axe less than 5, so that the frequncies for r =3,4,5 , 6 have been pooled together (see in table). No. of errors ( r )
Obseved frequency ( f , )
frequency ( / , )
f'
0
36 40 19
37 37 18
0.027 0.243 0.055
1 2
3 4
0
5
2
6
1 J
Total
2
6
1
iV* 1 0 0
1
1.125
2
[ 5
1 8
0
0
//=
100
J
X2 = 1.450
Step 4 : The tabulated value ofy* at 5%level of significance and fordofv-6 -1 =5 is 11.07 i.e., .*$.05, j —11.07.
4921Enumiuumg MxiwMxncs-fll
S b a i : Decision: Clearly calculated value of x2 = 1-45 < tabulated value of xl.os, 5=11.07. => the null hypothesis is accepted. => the errors are distributed according to Poisson probability law.
539 /-STATISTIC AND/-DISTRIBUTION: The statistic t was introduced by W.S. Gosset in 1908. Let x\7x2, jc* be the values of random sarnble of size n, drawn from a nonnal population with mean ji and variance a 2 (unknown). Then we define the /-statistic as : * where
-
S* = O T l)
(X'~ * y
1 £ Jc = sample mean - — Ex,. n i=l Here S is the standard deviation of sample with degrees of freedom v - n - 1, when n < 30. If we calculate t for each sample, we obtain the sample distribution for t. This distribution known as student's t-distribvtion is given by
and
....................... H ) Where y0 is a constant, so chosen that the total area under the curve is 1.
Remark: When random sample of size n is large, or [when population variance are known] that /-statistic define as:
where
a2 = ^
2
(xt - x Y -
S m sn c t | 493 5 .4 0 C O M P U T A T I O N O R W O R K I N G R U L E O F l J H S T R f f i U T l O N :
(a) To test the significance of a mean (snail sample) Step 1 : Consider the Null Hypothesis ffo: There is no Significant difference is between the sample means and population means. Alternative Hypothesis H% : There is significant difference in between the sample means and population means. Step 2 : Calcualte /-statistic : [when variance is not known] where
- (^ i)
When variance is known: then where
o 2 ** “fw I ( x - x ) 2. Step 3 : Level ofSignificance or Tabulated value of t : See the value of / from the table i.e.» value of t at a% level of significance and for the dof v. St£&4 : Decision: (i) If calculated value of |/j< tabulated value off. Then accept the null hypothesis H0. (ii) If calculated value of J/|Af tabulated value of t. Then reject the null hypothesis Ho. (b) To test the s&rificaace of dtffcreacc hetweea two sample weans : Let xu x2, xHi and y u yi* «•« y *2 be two independent random samples with means x and y and standard deviation S\ and $ from a normal population with the same variance. Suppose we have to test the hypothesis that the population means |ij and (i2 are the same, for it we calculate /-statistic as follows : 1
SSzH p* j22_ S
i ( n 1+n2) ’
S2 - ------- ---------- s----- ■* Hi + n2 - 2 with degrees of freedom v - n\ + t*i - 2 . When variances cr? and aj are known : Then 2 2 „ »i<*i + *2<*2 »1 + fl2 - 2 * Remarks : When variance b unknown : 1. For dof v and 95% confidence limits for p are given by : c J1 = 7 ± to.03 v>* “7*» where
"•'l>
4 9 4 1 E n o o c b b m q M athem atics -III
2. For dof v and 99% confidence limits for \i are given by: g M “ 5 ± /o.ei *» x "7“ Vit When variance is known : 1. For dof v and 95% confidence limits for p are given by :
<
|i * jp ± 4mh« x
4n- 1 2. For dof v and 99% confidence limits for ji are given by: i i ssx ± y i » x
a
Example 5.86 : Find the student's t-statisticfor thefollowing variable values in a sample : -4 ,-2 ,-2 ,6 ,2 ,2 ,3 ,3 , taking the mean of the universe to be zero. Solution : We have /-statistic : / Where
...d)
=
Z ( * -x )2
...(2)
Here n = 8 , and mean of universe p - 0. Serial No.
X
1
-4
2
-2
3 4 5
-2 2
6
2
7
3 3
8
x- X
( * - 3c)1
- 4.25 -2.25 -2.25 -0.25 1.75 1.75 2.75 2.75
18.0625 5.0625 5.0625 0.0625 3.0625 3.0625 7.5625 7.5625
Zx« 2
49.5000
vT o x - Mean - ~ = f =0.25 71 o and
S~
1 » -l
= J i p = 2.659 i 7
[by (2 )]
From (1), we have (0 2 5 -0 ) V8 . 2.659
Ans.
Svftnsnc* ( 4 9 9
Example 5.87 : A sampoo manufacturing company was distributed a particular brand of sampoc through a large number of retail shops. Before a heavy advertisement Campaign, the mean sales per sampoo was 140 dozens. After the campaign a sample of 26 sampoo was taken and the mean sales figure was found to be 147 dozens with standard deviation 16. Can you consider the advertisement effective 7 (Given /***2s m 1.708) Solution : Given: n - 26 (small samples) mean * y * 147; S.D. Le.» ct= 16 (variance are known), and degrees of feedom v *=n -1 * 26 - 1 * 25. StgflLl : Null hypothesis Hq : There is no difference in between the sample means and population means, i.e., the advertisement is not effective Le., H q : * 140. \A : Calculation of/-statistic : I We have
t = ——
—-
[when S.D. a is known]
_ (147-140)^26 ----------- 16----= 2.19. Step 3 : The tabulated value of / at 5% level of significance and for v * 25 dof is 1.708. i.e., fo.os, 25 = 1.708. Step 4 : Decision : Clearly calculate value of |f| - 2.19 K tabulated value of *0.03,25 = 1*708. => Null hypothesis Hqis rejected. => The advertisement is effective. Ans. Example 5.88 : Ten objects are chosen at randomfrom a population and their heigktsarefound to be in inches 63, 63, 64, 65, 66, 69, 69, 70, 70, 71. Discuss the suggestion that the mean height in the universe is 65 inches, given thatfor 9 dof the value o ft and 5% level of significance is Z262. Solution : Given n » 10, (small smaples) Universe mean : \i = 65 and variance is not known and degrees of freedom v * = n - l ” 9. Step 1 : Null hypothesis Hq: There is no difference between the mean height of sample and universe i.e., p » 65. Step 2 : Calculate of /-statistic : We have Where
t'
...(1)
71-1
...(2)
4 M | E n g b c h m g M athematics-IR
Zx mean x ~ n
and
670
67.
10
(x- x r
X
16 16 9 4
63 63 64 65
-4 -4 -3 - 2
66
- 1 ,
69 70 70 70 71
1
4 4 9 9 16
2 2
3 3 4
Z(x- jc)^ = 88
670 From die table, then equation (2) becomes *
9 = 3.13 inches
Hence (1) becomes t = ( 6 7 - g y i Q =2Q24 S s p j : The tabulated value of t at 5% level of significance and dof v = 9 is 2.262 i.e., ^0.05,9 = 2.262 Step 4 : Decision: Clearly calculated value of j/j312 .0 2 4 < tabulated value of to.05, 9 - 2 .262. => the null hypothesis Ho is accepted. => die mean height of the universe is 65 inches. Ans. Example 5.89: A salesman is expected to effect an average sales ofRs. 3500. A sample test revealed that asparticular salesman had made thefollowing sales. Rs. 3700,3400,2500,5200,3000and 2000. Conclude whether his work is belowstandard or not Using 5% level of significance of t and 5 dof is 2.015.
~
Solution : Given
S ta ti s tic s
j 497
n - 6 , (small sample)
Mean of sample:
3700+ 3400 + 2500 + 5200 + 3000 + 2000 J - ------------------------g------------------------
or
3c =
= 3300 and n - 3500
Here variance is not known and degrees of freedom v = w- 1=5. Step 1 : Null Hypothesis Ho : the salesman is up to standard. Step 2 : Calculation of /-statistic : We have
t = -—
Where
S2 *
...(1) ...(2 )
71—1
3700 3400 2500 5200 3000 2000
1* 1
&
x —jg
Sale in Rs. x
400
160000
100
10000
- 800 1900 -300 - 1300
640000 3610000 90000 1690000 S ( x - x >2 = 6200000
Ix = 19800 From the table, we have S1 => Hence (1) becomes:
= 1240000 o 5 = 1113.55 Rs.
[by (2)]
(3300 - 3800) s ’ na55 or / * - 0.44 | / 1 = 0.44 Step 3 : The tabulated value of t at 5% level of significance and dof v = 5 is 2.015 i.e., *0.05. 5 = 2.015. Step 4 ; Decision : Clarly calculate value of \ t | = 0.44 < tabulated value of *0.05, 5 = 2.015. => the hull hypothesis is accepted. => the salesman is upto standard. Ans.
4 9 8 | E n g in eerin g M athem atics -11 I
Example 5.90 : A random sample of size 16 has as mean is 53. The sum of the squares of the deviations taken from the mean is 1501 Can this sample be regarded as taken from the population having 56 as mean ? obtain 95% and 99% confidence Umits of the mean population. (For v * 15, taoi * 2.95 and tAB5 * 2.131). Solution: Given that n = 16 (small sample) Sample mean 3c = 53 and n = 56. Also
I (x - x ¥ - 150 and variance is not known. S2 =
n~l
= io. lb -l
=> S = y/lO => 5 = 3.162. The degrees of freedom v = n - J = 16 - 1 = 15. Step 1 : Null Hypothesis Hq : Population mean is 56 i.e., H0 : \i = 56. Step 2 : Calculation of /-statistic : We have,
/ =
...(1)
(53-56) Vl6 _ 3.162
=>
| / 1 - 3.79.
Step 3 : Given that tabulated values for dof v = 15 are :
(i) ^o.os, is = 2.131 (*0 *o.oi, t5 = 2.95. Sten 4 : Decision:
(i) Since calculated value of |/j = 3.79 K tabulated value of ^ 05,15 = 2.131 => the null hypothesis is rejected. => the population mean is not 56 at 5% level of significance. (ii) Since, calculated value of |/1 = 3.79 K tasulated value of fooi. 15 = 2.95 => the null hypothesis is rejected. => the population mean is not 56 at 1% level of significance. Again, we known that 95% confidence limits for ft (dof v = 15) are given by
S ta ti s tic s | 4 9 9
H = 5 3 ±2.131 x
H = 53 ± 1.684 50.316 < ^ < 54.684. 99% confidence limits for p (dof v = 15) are given by M“ x ± fo.oi x ~T= V/i = 53 ± 2.95 x 3.162 = 53 ±2.33, Ans.
50.67 < n < 55.33.
Example 5.91; A random sample of size 7from a normal population gave a mean o f977.51 and a standard deviation o f 4.42. Find 95% confidence interval for the population mean, (given that * - 2.447) n * 7, (small sample) Solution : Given that x = 977.51 and a = 4.42 (known) and dof v = « - 1=6. We known that 95% confidence limits for population mean ji for 6 dof. are given by
= 977.51 ± 2.447 x
442 V6
[*♦* fo.05,6 ~ 2.447]
= 977.51 ±4.416 973.09
Ans.
Example 5.92 : The nine items of a sample had thefollowing values: 45,47, 50, 52,48,47, 49, 53, 51. Does the mean of the nine items differ significantlyfrom the assumed population mean of 47.5 ? Given that
Where P is the area to the left of the ordinate at t
5 0 0 | E n g in eerin g M athem atics-111
Solution : and degrees of freedom JC 45 47 50 52 48 47 49 53 51
n - 9 (small sample) H - 47.5 and variance is not known v - n - 1 =8. x~ x
(x - x ) 2
-4.11 -2.11 0.89 2.89 -1.11 -2.11 -0.11 3.89 1.89
16.8921 4.4521 0.7921 8.3521 1.2321 4.4521 0.0121 15.1321 3.5721
Zx ~ 442
E(jc- x P = 54.8889 2* = 442 x = n = —9 =49.11
and
n -1
o
S = V6.8611 = 2.62 By /-statistic
= (x~\i)yfn S
_ (49.11-47.5) >/9 2.62 = 1.84. From the given data let us interpolate the value of P for / = 1.84 is as follows : fo r /=1.9 P = 0.953 / = 1.84 for / = 1.8 P = 0.945 /= 1.8 difference = 0.1
difference = 0.008
difference * 0.04
(Diff. in P) (Diff. in calculated value of t ) Difference -----------------Diff. of t as given--------------
S ta tistics | 5 0 1
Thus required value of P for P « 0.945 + 0.0032 * 0.9482 = 0.95 Thus *0.95 t - Area to the right of / = 1.84 = !-/> ,.i.m= 1 -0.95 = 0.05. Hence /’ (I/I* 1.84) - Left + Right = 0.05 + 0.05 = 0.1 which is greater than 0.05 | / 1 * 1.84 > 0.05 => Chance of getting value of t is greater than 0.05. The value of / is not significant => Sample may be a random sample which is drawn from the normal population with mean 47.5. Example 5.93 : The average number of articles produced by two machines per day are 200 and 250 with standard deviations 20 and 25 respectively on the basis ofrecords of 25 daysproduction. Can you regard both the machines equally efficient at 1% level of significance ? (Given that t6.0j,*s * 2.58) Volution : Given that fli » 25, xl = 200, S.D. C| = 20 n2 ~ 25, x2 ~ 250, S.D. a2 = 25 We have
p = "l°l + »2°rl _ 25 x (20)2 + 25 x (25)2 25 + 25 - 2
S = V533.85 = 23.1 Degrees of freedom v = n\ + n2 - 2 = 48. Step I : Null Hypothesis Ho : both the machines are equally efficient i.e., Hi = H2Step 2 : Calculate of /-statistic :
5 0 2 | E n g in e e rin g M a th e m a tic s-! 11
-50 23.1
x 3,54
- - 7.65 => | / 1= 7.65 Step 3 : The tabulated value of / at 1% level of significance and dof v = 48 is 2.58 i.e., *o.oi, 4* = 2.58. Step 4 : Decision : Since calculated value of |/| = 7,65 K tabulated value of fo.oi, 48 = 2.58. => the null hypothesis is rejected. => the two machines are not equally efficient at!% level of significance. Aas. Example 5.94: Two horses A and B were tested according to the time (in seconds) to run a particular track with the following results: HorseA 28 30 32 33 33 29 34 Horse B 29 30 30 24 27 29 Test whether you can discriminate between two horses♦ (Given that t&os, n “ 2.20). Solution : Given that «i = 7, n2 - 6 _
mean of A :
2 * _ 219
x ~ nx ~
31.29
7
169 28.17. 6 Degress of freedom v = «t + »2 - 2 = 11. Step 1 : Null Hypothesis Hq : No discriminate between two horses i.e., Ho: Hi = *i2. Step 2 : Calculation of /-statistic:
mean of B :
y = *2
t=
We have
x -y
€
n|»2
Table for calculation of S X
(x - x )
(x - X?
y
y- y
(y-y)2
28 30 32 33 33 29 34
-3.29 - 1.29 0.71 1.71 1.71 -2.29 2.71
10.8241 1.6641 0.5041 2.9241 2.9241 5.2441 7.3441
29 30 30 24 27 29
0.83 1.83 1.83 -4.17 - 1.17 0.83
0.6889 3.3489 3.3489 17.3889 1.3689 0.6889
31.4287
t y = 169
Lx - 219
;
26.8334
S tatistics |
Since
503
Z ( x - * ) 2 +Z ( y - y ) 2
„
n l + n 2 ~
2
3L4287 + 26.8334 7+6 -2
S = VS!296 = 2.30 Hence (1) becomes /=
(31.29-28.17) '7x6 7+6 2.30 .
5.608 • 2.30 = 2.44. Step 3 : The calculated value of t at 5%level of significance and for dof v = 11 is 2.20 i.e., *0.05, n = 2.20. Step 4 : Decision : Since the calculated value of |/| 2.44 < tabulated value of t = 2.20. => the null hypothesis is rejected => there are dicriminate between two horses at 5% level of significance. Ans. xample 5.95 : Below are given the gain in weights in kgs of cowsfed on two diets x andy. Diet x : 13 14 10 11 12 16 10 8 11 12 9 12 Diety: 7 11 10 8 10 13 9. Test at 5% level of significance whether the two diets differ as regards their effect on mean increase in weight (given that n «*2.11) olution : Given that: n\ ~ 12, n2^ l and degrees of freedom v = nx+ n2 - 7 = 17. =
bT
= 1 2 ' = 11S
Step 1 : Null Hypothesis H0 : two diets do not differ significantly i.e., Ho : Hi: Step 2 : Calculation of /-statistic :
504
| E ngin eerin g M a ih em a h c s -III
Table for calculation o f S : X
x -
13 14 10 11 12 16 10 8 11 12 9 12
x
(x -x ?
2.25 6.25 2.25 0.25 0.25 20.25 2.25 12.25 0.25 0.25 6.25 0.25
1.5 2.5 - 1.5 -0.5 0.5 4.5 -1.5 -3.5 -0.5 0.5 -2 .5 0.5
( y -y ) 2
7 11 10 8 10 13 9
-2.71 1.29 0.29 - 1.71 0.29 3.29 -0.71
7.3441 1.6641 0.0841 2.9241 0.0841 10.8241 0.5041
Sy = 68
23.4256
S ( * - i ) 2 + S (y -J )' -------n ^ n t - 2
_
Smce
y-y
t
53.00
138
y
*
5a00+2a4256 12 + 7 - 2 « 4.4956
S = V4.4956 =
2.12
From (1), we have 212
>.71) 112x7 V12 + 7
- 1.77. Step 3 : The tabulated value of *at 5%level of significance and for dof v = 15 is 2.11 i.e., *0.05.15 = 2.11 Step 4 : Decision : Since calculated value |*| = 1.77 < tabulated value o f *0.05, 15 = 2.11. => null hypothesis is accepted. => the two diets do not differ significantly. Ans.
S tatistics
| 505
5.41 FISHER’S F-TEST The F-test was first originated by a great statistician even agriculturist prof. R.A. fisher. The test is also known as Fisher’s F-test or simply F-test It is based on F-districution. The F-test refers to a test of hypothesis concerning two variances derived from two samples. The distribution of variance ratio F with vt and v2 degres of freedom is given by :
where y0 is so chosen that the total area under the curve is unity.
5.42 F-STATISTIC: Let xu *2, ....* x„u and y x%y i , ....y*i be the values of two independent random samples drawn from the two normal populations with mean iii and n2 and variances ai2, a22 respectively, then we define variance ratio F as :
when
and jc, y are the sample means, with vi = n\ - 1, and v2 = Note : The value of F-statistic is always greater then L
- 1 are degrees of freedom.
5.43 ASSUMPTIONS IN F-TEST The F-test is based on the following assumptions : (i) The values in each group should be normally distributed. (ii) The error should be independent of each value. (iii) The variances within each group should be equal for all groups. 5.44 COMPUTATION OF F-TEST: We have the following steps : Step 1 : Null hypothesis Ho : ctj2 = a22 Alternative Hypothesis H\ :
5 0 6 | E n g in e e rin g M a th e m a tic s-! 11 S 2
or
F = — , when S22 £ Si2. Sf
Step 3 : Choose tabulated value of / ’-statistic at a% level of significant with degrees of freedoms vj and V2. Step 4 : Decision : If calculated value of F < tabulated value of F^*\, v2) Then accept null hypothesis Hq. If calculated value F \ tabulated value of i(vi, V2> Then reject null hypothesis Hq. Example 5.96 : Random samples are drawn from two populations and the following results were obtained: Sample X : 20 16 26 27 23 22 18 24 25 19 Sample Y: 27 33 42 35 32 34 38 28 41 43 39 37 \ Find variance of two populations and test whether the two samples have same variance. (Given that F&0Sfor 11 and 9 dof is 3.112). Solution : Given that wj = 10 and n2 - 12 with degrees of freedom v = n\ - 1 = 9 and V=
- 1 — 11.
Step 1 : Null Hybothesis Hq : Let CTj2 = 022 i.e., the two samples have the same variance. Step 2 : Calculation of F-statistic : We have to find Si2 and S22 : Sample X y
-2 -6 4 5 1 0 -4 2 3 -3
4 36 16 25 1 0 16 4 9 9
27 33 42 35 32 34 38 28 41 43 30 37
0
120
y -y
(y-y)2
-8
6 8 -5 2
64 4 49 0 9 1 9 49 36 64 25 4
0
314
-2 7 0 -3
-
1 3
-7
0
Ex = 220
(x-x)2
tl -f* to.
20 16 26 27 23 22 -----18 24 25 19
x- x
V:
X
Sample Y
I S ta tistics
Here
n\ = 10, -
=
wj - 12
420 - . a = 35 12 ^ »2 314 £ (y -y )* = 28.5 11 *2 - l
« 220 =
x
nx 10 SQc - x)2 Si2 = fli~ 1 Hence F
I 507
= £2_= gas Si2 " 103
22* 120
= 133
2.14
[V
S22 > S ,2 ]
Hence calculated value of Fis 2.14. Step 3 : The tabulated value of F at 5% level of significance for the dof v = 11 and 9 is 3.112 i.e.t Fo.os - 3.112. Step 4 : Decision : Calculated value of F = 2.14 < tabulated value of F0.05 = 3.112 => the null hypothesis H0 is accepted. => the two samples have the same variance. Example 5.97: in a test given two groups o f students drawtffrom two normalpopulations, the marks obtained were asfollows: Group A : 18 20 36 50 49 36 34 49 41 Group B : 29 28 26 35 30 44 46 Examine at 5% level, whether the two populations have the same variance, (Given that Fe.83(*,6) “ 4,15). Solution : Given that: n\ = 9, and n2 = 7
and
jjc * “ »i
333 9
-
238
SL
37 34.
Step 1 : Null Hypothesis H%: ctj2 = a^, Le., two populations have the same variance. Step 2 : Calculation of F-statistic: Group B
GroupA x 18 20 36 50 49 36 34 49 41 Ex = 333
x -x -1 9 -1 7 -1 13 12 - 1 -3 12 4
(x-x)* 361 289 1 169 144 1 9 144 16 1134
y 29 28 26 35 30 44 46
= 238
y-y -5
-6 -8 1 -4 10 12
(y-y? 25 36 64 1 16 100 144
386
5 0 8 | E ngin eerin g M athem atics-111
We have
Z ( x - x ) 2 1134 . _ Si2 ---------nj -1;— = —8Q— = 141.75
Clearly
z(y-y)2 n2 - 1 Si2 > S22 Si2 =
238 6 = 64.33
S{ 141.75 F = c2 64.33 *^2 i.e., F “ 2.203 Step 3 : The tabulated value of Fat 5% level of significance and for degress of freedom 8 and 6 is 4.15, i.e., Fqm = 4.15 Step 4 : Decision : Since calculated value of F ~ 2.203 < tabulated value of Fo.os = 415. => the null hypothesis Ho is accepted. => the two population have the same variance. ( Ans. Example 5.98 : In a laboratory experiment, two samples gave thefollowing results: 111611
Sample
Size
Sample mean
Sum o f squares of deviation from the mean
I
10
15
90
i
12
14
108
Test the equality ofsample variance at 5% level ofsignificance. (Given that />, n (U.os) - 2.90) Solution : Given that: = 10 n2 = 12 jc
y - 14
~ 15
Z ( y - y)2 = 108 Z { x - xY =90 Step 1 : Null hypothesis Ho . C\2 = o22 i.e., two samples have the same variance. Step 2 : Calculated of F-statistic : '
We have and Clearly
£ ( s - s-?\2 ) 5,2 = Hi -1
90 9
=
10
108 z(y-y)2 = 9.82 n 11 n2 - 1S22, then c2 10 = 1.018 f ~ -i-*22 =
sS
982
Step 3 : The tabulated value of F at 5% level of significance for dof 9 and 11 is 2.90 i.e., Foos = 2.90.
S ta tistics |
509
Step 4 : Decision : Since calcualted value of F ~ 1.018 < tabulated value of Fo.os ~ 2 .9 0 .
the null hypothesis //<> is accepted. the two populations have the same variance.
Ans.
5 . 4 5 F I S H E R ’S z - D I S T R I B U T l O N
l^t x \ , x 2, ...... .v„i and y i% y2.......yn2 be the values of two independent random samples'with estimated variance S{2 and S22. Then Fisher’s /-distribution is the statistical distribution of halfof the logarithm of an F-distribution variate: z ^ |l o g « F
i.e .,
S1 = 2o log. ^ S^V ,I; where S\2 > S22. With degress of freedom vt ~ n\ - 1 and i>2 = n2 - I. It was first described by Ronald Fisher in 1924. The probability density function of /-distribution is defined as follows : u\!2 y
-
------------- --------- -- X --------------------------- -.----------- r j ~ ; - o o < z < OC
a t e .* )
Remark: With the help of equation (1), all the applications of F-distribtution may be regarded as the application of z-distribution also. Example 5.99 : Two gauge operators are testedfor precision in making measurement One operator completes a set of26 readings with a standard deviation of 134 and the other does 34 reading with a standard deviation of 0.9%. What is the level ofsignificance of this difference ? Given that for dof v, * 25 and v2 m33, the value of z*os * 0.306 and z+oi *=0.432. Solution : Given . = 26 and n2 ~ 34 S.D. at = 1.34 and o2 = 0.98 Since => => =>
, Z(x~x)2 , U y~ y)2 Oj2 - ---------- and o22 - ----“----rt, n2 (1-34)2 -
and (0.98F =
I ( x - x f - (1.34)2 x 26 and 2. ( y - y Y = (0.98)2 * 34 I. (x- x)2 ~ 46.68 and I iy - y? = 32.6536
5 1 0 | E n g m e h b n g M athem atics-111
Step I : Null Hypothesis H0 : The difference between variances is not significant i.e., O!2 =
and Clearly
Si2 >
we have
= | log, (1.8770) = 0.3148. Step 3 : The tabulated value z at 5%and i % level of singificance and for dof 25 and 33 are zoos —0.306 and zo.ot36 0.432. Step 4 : Desicion : (1) Since Calculated value of z = 0.3148 K tabulated value of £o.os= 3.06 => the null hypothesis Ho is rejected. => the difference between variances is significant at 5% level of significance. (ii) Since calculated value of z = 0.3148 < tabulated value of zo.oi ~ 0.432. => the null hypothesis H0 is accepted. => the difference between variances is not significant at 1%level of significance. Ans. Example 5.100 : Test whether the two sets of observations: I : 17 27 18 25 27 29 27 23 17 n : 16 16 20 16 20 17 15 21 Indicate samples drawnfrom the same universe. (The value o f z at 5% levelfor 8 and 7 degrees o f freedom Is 0.6575J. Solution : Given that /i, =9 J»2 = 8 with dof V| = «2 - 1 = 8 and v2 = *2 - 1 = 7.
S ta tistics |
5 ll
Step 1 : Null hypothesis Ho : Two population variances are same i.e., H0 : ai2 = O22. Calculation of z-statistic: 11 observation
I observation X
x- X
17 27 18 25 27 29 27 23 17
-6.33 3.67 -5.33 1.67 3.67 5.67 3.67 -0.33 -6.33
y
y-y
40.0689 13.4689 28.4089 2.7889 13.4689 32.1489 13.4689 0.1089 40.0689
16 16 20 16 20 17 15 21
- 1.625 - 1.625 2.375 - 1.625 2.375 - 0.625 - 2.625 3.375
184.6001
141
(x- xY
210
0>- y Y 2.6406 . 2.6406 5.6406 2.6406 5.6406 0.3906 6.8906 11.3906 37.8748
We have _
S'
_ 184.0001 = , ,
£ (*-3i f
n i- 1
8
23
w . a t B i . 2 z * 6 a - M107 n2 - 1 Since
S\2 > S22 ,2 r l ^i 5
z = i lo&
2
= 1 10& (sfiW ) = \ log, (4.2508) = 0.7236. Step 3 : The tabulated value of z at 5% level of significance with dof 8 and 7 is 0.6575. i.e., 20.05 = 0.6575. Step 4 : Decision: Since calculated value of z = 0.7236 K tabulated value of zoos= 0.6575. => the null hypothesis is rejected. => the two variances are not same. Ans.
5 J & I E n q w e e o n g M athemattcs-III
Exercise-5 (A) 1.
Find whether the following function is a p d.f. r f x _ J *» 0 < ; x s l J W “ |2x, 1 £ x £ 2 ’
2.
A continuous random variable A"has />.<£/ /(x ) = 3x2»0 ^x ^ 1 Find the value of ‘a \ if P(X<>a) = P ( X > a). 0, x <2 •^■(3 + 2x), 2 < x < 4 8 0, *>4
3.
If
4.
Prove that/ (x) is a p .d f Also compute P (2 < X < 3). A random variable X has the p.d.f.
f ix ) -
x,
0 £ x <1
f i x ) _ •{2 - x , 1 £ jc < 2 0, x 2: 2
Find distributive function or cumulative distributive function (c.d.f.) of X. A random variable X has p.df. :
/far3*' *>o jo , 6.
is o
Find (i) the value of k (ii) compute P (1 < X < 2) and P (X> 3). A random variable A'has a p.d.f. x 01.5).
S ta tistics ) 5 1 3
7.
Find the mean o f the follow ing p .d .f.
fix)
8.
0,
otherwise.
For the continuous distribution : dF = y0 (x - jc2) dx, 0 £ x <, 1. Find
9.
mean, median and mode.
If a random variable X has p.df. fix) = ^
-aZxZa.
Find first, second, third and fourth moments about origin. 10. A p.df. function defined by — (3 + x)2, - 3 < * £ --! fix ) =
X
( 6 - 2 * 2 ),
~ ( 3 - a:)2,
- 1 <
x
< 1
1 < x £ 3.
Find mean, variance and standard deviation. 11.
1
Vox ^ p.d f f {x) -
Find mean, median, and mode. 12. A distributive function F (x) is defined as 0
, x <>2
1
x >£
F (x)-
Find p.df. of x. Also find mean.
Answers-5 (A) 1. 2. 3.
* 7 No., because J f(x) dx = « * 1. a * (0.5)l/3. 4/9
514 \ E n g in e e rin g M a th e m a ttc s-III
fo
, x<0
4.
F(x) = 2* 1
5.
k -3 ,
6.
I 3 15 8 ’ 8’ 32 ‘
2
l * x<2 , xZ 2
P ( I < A*< 2) = e*3 - e** and /> (A' £ 3) = e~9.
13 5 8.
_y0 = 6 , mean =
median =
mode = ^
9.
m' = o,
10. 11.
mean = 0, variance = 1, S.D. = 1. mean = mode = median - 0 .
^ ,M 3 ’ = o, h4'=
,_ x+3 12. a = , mean = 5.375
Exercise-5 (B) For a p.d.f f(x) /(* )=
s - G <* <6 [ 0, otherwise Find the value of k and show that/ (x) is a rectangular distribution. Also fmd mean. For the rectangular distribution ; f i x ) ~ fjj~ , - a < x < a
Find \i2, ^3 and m (i.e., 2nd, 3rd & 4thmoments about mean). For the rectangular distrubiton : f(x\ = \ 2k [ 0,
10-fe
otherwise
Find mean and variance. A lso find distribution function
F (x).
S ta tistics I 5 1 5
4.
Given the Incomplete Beta function: x
3 (I fn)
B, (/, m) * J jc^1(1 - *)"-1dx and lx (/, m) = 5.
then prove that Ix f/, tti) * 1 - /lHr (m, /) The income tax of a man is exponentially distributed with the p.d.f is given by 1 --x/3 /<*) = 3* What is the probability that his income will exceed Rs. 17000 assuning that the income tax is levied at the rate of 15% on the income whose Rs. 15,000 ? The life time X in hours of a T.V. tube of a certain type obeys an exponential distribution with X = 0.001 hours. Find (i) P (X> 1000), (ii) P (700 ZX<> 100) Find the value of *c\ so that / ( jc ) be a p.df. of/ ( jc ) = ce-2*, x > 0, then prove that (i) fiQ O M(Jffc), (ii) Cwfficient of variation is 1.
Answers-5 (B) k - b ~ a, mean = ^ (a + b). 2
4
|«2 ^ "jp
® IT*
mean = 10, variance =
and
jc- ( 1 0 - £)
d.F.
F(x) =
10-/t
xZ 10+£. -100 6.
(i) 0.368, (ii) 0.129. j* **A f* |»
m
>
m
Exercise-5 (C) 1. 2. 3.
Find the Binomial distribution whose mean is 5 and variance is 10/3. Find the Binomial distribution whose sum of mean and variance is 4.8 for 5 trials. In a Binomial distribution, the sum of the mean and variance for 5 trials is 1.8. Determine the distribution.
516 | Engineering Mathematics-III
4.
5. 6.
If 10% of the rivets produced by a machine are defective, find the probability that out of 5 rivets chosen at random : (i) None will be defective, (ii) One will be defective, (iii) At least two will be defective. If the probability that a newborn child is a male is 0.6, find the probability that in a family of 5 children there are exactly 3 boys. Three percent of a given lot of manufactured parts are defective. What is the probability that in a sample of four items none will be defective. [Hint.
7.
/>(0) =4 C0(O.O3)°(O.97)4 = 0.885 ]
The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 pass in the examination. [Hint.
p= 0A,
<7=0.6
/»(*;>4)=P(x = 4)+/>(jc=5)+P(jc=6). =C4(.6)4(.4)2 +C5(0.6)5(0.4)=C6 (0.6)6=0.5443]. 8.
Five coins are tossed 3,200 times. Find the frequencies of the distribution of heads and tails and tabulate the result. Calculate the mean number of successes and standard deviation. [Hint.
9.
In a particular market 40% of the consumers prefer readymade clothing. A sample of n = 5 consumers is to be drawn. Give the probabilities of having 0,1,2,3,4 and 5 preferring readymade clothes. [Hint. Probability of r consumers preferring readymade clothes out of a sample of 5=5CX(0.4)r(0.6)5_r. Put r =0,1,2,3,4 and 5.]
10. The incidence of occupational disease in an industry is such that the workers have a 20% chance of suffering from it. What is the probability that out of six workers chosen at random four or more will suffer from the disease. [Hint.
Here p=0.2, ^=0.8, n=6. p(r ;>4)=C4(0.2)4(0.8)2 +C5(0.2)5(0.8)+C6(0.2)6].
S ta ti s tic s | 517
11.
A sample of 10 pieces was examined out of a large consignment which has 5% defective pieces. Give the probability of 1 defective in the sample of 10. [Hint Here
12.
10 0.05, ?=0.95, n=10 />(r= 1)=C,(0.05)(0.95)9].
In 100 sets of ten tosses of an unbiased coin, in how many cases should we expect seven heads and three tails. [Hint Here Probability of head
»$=—,»=10 #=100.
ioo— Vj rI —I n 3' Required number of cases »100xC7l 13.
Fit a Binomial distribution for the following data and compare the theoretical frequencies with the actual ones: x : 0 I 2 3 4 5 / : 2 14 i0 34 22 8 14. In a Binomial distribution, for n = 5, if P(r - 1) = 0.4096 and P(r - 2 ) - 0.2048, then find the value of p. 15. Ten coins are tossed 1024 times and the following frequencies observed. Compare these frequencies with the expected frequencies. x : 0 I 2 3 4 5 6 7 8 9 10 / : 2 10 38 106 188 257 226 128 59 7 3 16. Seven coins are tossed and the number of heads are noted. This experiment is repeated 128 times and the following distribution is obtained: x : 0 1 2 3 4 5 6 7 Total /: 7 6 19 35 30 23 7 1 128 Fit the binomial distribution considering the coin is unbiased. 17. Fit a Binomial distribution for the following data and find the expected frequencies of the chance of machine being defective is 1/2. x : 0 1 2 3 4 5 6 7 / : 7 6 19 35 30 23 7 1
Answers-5 (C)
518 | E n g in e e rin g M a th e m a ttc s-III
4. 5.
(i) 0.5905, 0.3456.
9-
C,(0.6)r(0.4)5*', r = 0,1,2,3,4,5.
1°*
(ii) 0.3281, 6. 0.885.
(iii) 0.0815. 7.0.5443.
8. Mean = 2.5, S.D. =1.118.
C a { 0.2)4 + C5(0.2)5(0.8)+(0.2)6.
/ ] V° 12. i°0xC7|j J 10
11.
,0C, (0.05)(0.95)9.
13.
/(r) = 100x5 Cr(0.432)5~f(0.668)r, r - 0, I, 2, 3, 4, 5.
14.
I P=?
15.
1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1.
16.
128(0.517+ 0.423)7; 1, 8, 23, 36, 33, 19, 6, I
17.
1 ,7 ,21,2,5,21,7,1.
Exercise-5 (D) 1.
2.
If a random variable has a Poisson distribution such that P (1) -P (2), find (i) Mean of the distribution (ii) P{4). 2 Suppose that X has a Poisson distribution. If P{X - 2) ~~P{X = I), find, (i) P(X = 0)
3. 4.
5.
(ii) A ^ = 3).
A certain screw making machine produces on average 2 defective screws out of 100, and packs them in boxes of 500. Find the probability that a box contains 15 defective screws. The probability that a man aged 35 years will die before reaching the age of 40 years may be taken as 0.018. Out of a group of 400 men, now aged 35 years, what is the probability that 2 men will die within the next 5 years ? Suppose a book of 585 pages contains 43 typographical errors. If these errors are randomly distributed throughout the book, what is the probability that 10 pages, selected at random, will be free from errors ?
S ta tistics |
519
Using Poisson distribution, find the probability that the ace of spades will be drawn from a pack of well-shuffled cards at least once in 104 consecutive trials. If m and prdenote by the mean and central rth moment of a Poisson distribution, then prove that am In a normal summer, a truck driver gets on an average one puncture in 1000 km. Applying Poisson distribution, find the probability that he will havie *. (i) No puncture (ii) Two punctiires in a journey of 3000 kms. j Six coins are tossed 6400 times. Using Poisson's distribution find approximate probability of getting 6 reads x times. A manufacturer knows that the razor blades he makes contain on an average 0.5% of defectives. He packs them in packets of 5. What is the probability that a packet picked at'random will contain 3 or more faulty blades ? A source of water is known to contain bacteria with mean number of bacteria per cc equal to 2. Five 1 cc test tubes were filled with water. Assuming that Poisson distribution is applicable, calculate the probability that exactly 2 test tube£ contain at least* 1 bacterium each. Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in a sample of 10 tools chosen at random (i) Exactly two will be defective and (ii) More than one will be defective, by using Poisson distribution. Between the hours 2 P.M. and 4 P.M., the average number of phone calls per minute coming into the switchboard of a company is 2.35. Find the probability that during one particular minute there will be at most 2 phone calls. [Hint
n/ , e mm' e~235 x(2.35)r p(r) = — T ’= ----------------- » r = °»X' 2»3*
P(X 2) = P(X = 0)+P(X = 1) + P(X = 2), =
, 2.35 (2.35)'v2 1+
-------- * +
1!
2!
= 0.583].
If 5%of the electric bulbs manufactured by a company are defective, use Poisson distribution to find the probability that in a sample of 100 bulbs, 5 bulbs will be defective. [Hint
5 2 0 | E n g in eerin g M athem atics-!!!
15.
16. 17.
18.
19.
20.
21.
The number of accidents in a year attributed to taxi drivers in a city follows Poisson distribution with mean 3. Out of !,000 taxi drivers, find approximately the number of drivers with (i) no accidents in a year, and (ii) more than 3 accidents m a year. The probability that a man aged 50 years will die within a year is 0.01125. What is the probability that out of 12 such men at least 11 wilt reach their fifty-first birthday. A machine produces large quantities of hems and past experiment shows that on an average it produces 1%defective items. A sample of 20 items in drawn from a large number of Hems. Use Poisson distribution to determine the probability of 2 defectives. Fit a Poisson distribution to the following data and calculate theoretical frequencies : Deaths: 0 1 2 3 4 5 6 7 Frequencies: 305 365 210 80 28 9 2 1 Fit a Poisson's distribution for the following data: x: 0 1 2 3 4 5 6 7 8 9 10 y: 103 143 98 42 8 4 2 0 0 0 0 From records of Prussian Army corps kept over 20 years, the following data was obtained showing the number of deaths caused by the kicks of a horse. Calculate the theoretical Poisson frequencies : Number ofdeaths 0 1 2 3 4 Tbtal Frequencies 109 65 22 3 1 200 Fit a Poisson's distribution for the following data: x: 0 1 2 3 4 5 Tbtal f: 229 211 93 35 7 1 576
Answers-5(D) 1.
(i)2
(ii) 2/32
2. 3.
(i) e-4 0.035.
(ii) 4e~*. 4. 0.01936.
8.
(i)
(ii) (415)
11. 12.
0.3459. (i) 0.18395
(ii) 0.2642,
5. 0.4795.
6. 0.865.
13. 0.583.
14. 0.1823.
(ii) 353 nearly 15. (i) lOOxe-3 = 49.8, 17. 0.0164. 18. 301.2, 361.4, 216.8. 86.7. 26.0, 6.2, 1.2, 0.2. 19. 107, 141, 93, 41, 14, 4, 1, 0, 0, 0, 0. 20. n = 109, 66, 20, 4, 1. 21. 227.26, 211.35, 98.28, 30.46, 7.08, 1.31.
16. 0.9916.
S tattstkb i 5 2 1
5.
Ex Find the probability that the standard nonnal variate lies between 0 to I 5. Find the area under the nonnal curve for (a) Z ~ 1.64, (b) Z *• 1 32. (c) Z ~ 0.56 (d) Z - J 54. Find the area to the (a) right of Z *■0.25; (b) left of Z * 1.96; (c) left of Z = - 1.96; (d) between Z = 0.4 and Z * 0.6. A large number of measurements in normally distributed with mean of 65.5 cm and a standard deviation of 6.2 cot. Find the percentage of measurements that fad between 54.8 cm. and 68.8 cm. In an intelligence test administered to 1,000 students the average score was 42 and standard deviation 24. Find
=*0.333.
P(Z > .333) = 0.5-0 1304- 0.3696. Expected number of students exceeding a score of 50 = .3696 x 1000 = 369.6*370. [Hint (b) WhenXhes between 30 and 54, then Z lies between - 0.5 and 0.5. Probability of having students with score between 30 and 54 = 0.1915 + 0.1915 - 0.3830. Expected number of students to score between 30 and 54 = 1.000x0.3830 = 383. 100 1,000
[Hint (c) Probability of getting top 100 students ------- = 0.1. Standard normal variate having 0.1 area to the right = 1.28. X -42 24
Also 6.
>X = 72.72 « 73
In a distribution exactly 7% of the items are 35 and 89% are under 63. What are the mean and standard deviations of the distribution ? [Hint
! 4 8 ^ 3 5 -n = I.48a
a
... (0
/ 'j
Z =
= j .23 ^
a
63 - n = -1.23o
Solve (1) and (2) we get ji = 50.3, o = 10.33 ].
t
.. (2)
5 2 2 | E n g in eerin g M athem /u tc s -III
7.
A sample of 100 diy battery cells tested to find the length of life produced the following results: |A=12hours, a = 3hours. Assuming the data to be normally distributed, what percentage of battery cells are expected to have life (a) more than 15 hours, (b) less than 6 hours (c) between 10 and 14 hours. [Hint (a)WhenX= IS, Z = ^ j^ * = l Area to the right of Z = 1 is 0.5-0.3413 = 0.1587. Percent of battery cells having life more than 15 hours = 0.1587 x 100 = 15.87%. [Hint (b) When X = 6, Z =
= -2
Area to the left of Z is -2 = 0.5 - 0.4772 = 0.0228. %of Battery cells having life less than 6 hours = 0.0228 x 100 = 2.28%. [Hint (c) When X = 10, Z = —~
=-0.67 When JIT=14, Z = -
2 =0.67
Area betweenX - 10 to X ~ 14 is twice the area to the left of Z = .67 = 2x0.2487 = 0.4974. 8.
% of Battery cells having life span between 10 hours and 14 hours = 49.74%]. The mean lifetime of 100 Watt light bulbs produced by Larsen and Turbo is 200 hours. It is known that the standard deviation is 20 hours. Assuming that the life time of light bulbs are normally distributed, what are the probabilities that a single 100 Watt light bulb extracted fromthe production lot will (a) bum out between 180 hours and 210 hours ? (b) bum out for a time greater than 250 hours. [Hint (a)WhenX= 100,then Z = - —
20
°^ = -land vhen*= 210 then Z = 2-
°
20
= 0.5
Area between Z= - 1 and z * 0.5 is 0.3413 + 0.1915 = 0.5328. Required probability = 0.5328. [Hint (b) When X = 250, Z = — - - = 2.5.
9.
Required Probability= Area to the right of Z equals 2.5 = (0.5 - 0.4938) = .0062] In a sample of 1,000 the mean weight is 45 kgs with standard deviation 15 kgs. Assuming the normality of the distribution, find the number of items weighing between 40 and 60 kgs.
= /X-0.333 <; Z £ 1) = P{r0.333 < Z < 0) + P(0 < Z < 1) = 0.1293 + 0.3413 = 0.4706. Required number of items = 1,000 x 0.4706 = 470.6 » 471].
S ta tis tic s | 5 2 0
10.
In a sample of 120 workers in a factory the mean and standard deviation of wages were Rs. 11.35 and Rs. 3.03 respectively. Find the percentage of workers getting wages between Rs. 9 and Rs. 17 in die whole factory assuming that the wages are normally distributed. 11. Fit the equation of the best fitting normal curve to the following distribution: x : 0 \ 2 3 4 5 / : 13 23 34 15 11 4 12. Fit a nonnal distributions to the following data: Mid point of interval 100 95 90 85 80 75 70 65 60 55 50 45 Frequency 0 1 3 2 7 12 10 9 5 3 2 0 13. Fit a normal distribution to the following data: Variable 60-62 63-65 66-68 69-71 72-74 Frequency 5 18 42 27 8
Answers-5(E) 1. 2. 3. 4. 6. 7. 8.
0.4332 (a) 0.4484;
II.
„ , ioo -(it)1 y= f(x )= -rrre^ V3.4*
12. 13.
(d) 0.4382. (d) 03811. (c) 73.
10. 75.1
Mean (*i) = 71.2, S.D. (a) = 9.95; Theoretical frequencies : 0.2, 0.6, 1.8, 4.1, 7.3, 10.1, 10.7, 8.9, 5.7, 2.9, 1.1, 0.3. Mean (p) - 67.5, S.D. (cr) = 2.9; Expected frequencies : 2.55, 13.43, 13.10, 18.64, 4.81. VMftx mi
f. * *r
>- X. '
Ml
«
h*
*,
'
Exercise-5 (F) 1.
A sample of 300 students of Under-Graduate and 300 students of Post-Graduate classes of a University were asked to give their opinion toward the autonomous colleges. 190 of the UnderGraduate and 210 of the Post-Graduate students favoured die autonomous status. Present the above data in the form of a frequency table and test at 5% levels, the opinions of Under-Graduate and Post-Graduate students on autonomous status of colleges are independent (Table value of chi-square at 5% level for 1 d.f. is 3.84).
5 2 4 | E ngin eerin g M athem atics -III
2.
3.
4.
5.
6.
7.
A certain drug is claimed to be effective in curing colds. In an experiment on 164 poeple with cold, half of them were given the drug and half of them given sugar pills. The patient’s reactions to the treatment are recorded in the following table. Test the hypothesis that the drug is not better than sugar pills for curing colds. Helped Harmed No effect Drug: 52 10 20 Sugar Pills: 44 12 26 (Table value of chi-square at 5% level for d f 2 is 5.99) In a survey of 200 boys, of which 75 were intelligent, 40 had skilled fathers, while 85 of the unintelligent boys had unskilled fathers. Do these support the hypothesis that skilled fathers have intelligent boys. Use Chi-square test. (Values of x2 for 1-degree of freedom at 5% level is 3.84). Four dice were thrown 112 times and the number of times 1, 3 or 5 were as under : Number of dice showing 1,3, or 5: 0 1 2 3 4 Frequency: 10 25 40 30 7 (Table value of chi-square at 5% level for 4 d.f. is 9.488). To test the efficiency of a new drug a controlled experiment was conducted where in 300 patients were administered the new drug and 200 other patients were not given the drug. The patients were monitored and results were obtained as follows : Total Condition No Cured effect worsens 60 300 Given the drug: 200 40 200 50 Not given the drug: 120 30 500 no Total 320 70 Use x2 test for finding the effect of the drug. (x0o$, 2 = 5.99) The following table gives the classification of 100 workers according to sex and the nature of work. Test whether the nature of work is independent of the sex of the work. Skilled Unskilled Male 40 20 Female 10 30 (Use xi.05.4 * 3.84) There is a general belief that high income families send their children to public schools and low income families send their children to Government schools. For this 1000 families were selected in a city and the following results were obtained. Income Public Schools Govt Schools Total Low 100 200 300 High 500 200 700 Total 600 400 1000 Use Chi-square test to determine whether income level and the type of schooling were associated. (Xoos. 1 = 3.84).
S ta tistics J 5 2 5
8.
Calculate chi-square for the follow ing data :
Class :
9.
10.
11.
12.
13.
14.
15.
A B C D Total 8 29 44 15 4 10.0 Jn : 7 24 38 24 7 100 f' Calculate the value of t in the case of two characters A and B whose corresponding values are given below: A 16 10 8 9 9 8 B 8 4 5 9 12 4 The table below are for protein tests of the same variety of wheat grown in two districts. The average in District I is 12.75 and in District II is 13.03. Calcualte / for testing the significance between the means of the two districts : Protein results District 1 12.6 13.4 11.9 12.8 13 District II 13.1 13.4 12.8 13.5 13.3 12.7 12.4 [Given that /oos. to ~ 2.228] Ten cartons are taken at random from an automatic filling machine. The mean net weight of the 10 cartons is 11.8 kg and standard deviation is 0.15 kg. Does the sample mean differ significantly from the intented weight of 12 kg ? (Given that for /0.05, - 2.26). A fertiliser mixing machine is set to give i 2 kg ofnitratefor every quintal bag of fertiliser. Ten, 100 kg bags are examined. The percentages of nitrate are as follows : II, 14, 13, 12. 13. 12, 13, 14, II, 12. Is there reason to believe that the machine is defective ? Value of t for 9 degrees of freedom is 2.2621. Ten students are selected at random from a college and their heights are found to be 100, 104, 108,110,118,120,122,12-4,126 and 128 cms. In the height of these data, discuss the suggestion that the mean height of the students of the college is 110 cms. (Given /005.9 = 2.26) Talcum powder is packed into tins by a machine. A random sample of 11 tins is drawn and their contents are found to weight in kgs as follows : 0.44. 0.51, 0.49, 0.52, 0.45, 0.48, 0.46, 0.45, 0.47, 0.45 and 0.47 Test if the avarage packing can be taken to be 0.5 kgs. (/0 05. 10 = 2.28) Samples of the types of electric bulbs were tested for the length of and the following data were obtained Type I Type II Sample No. 8 7 Sample Mean 1234 hours 1036 hours Sample S.D. 36 hours 40 hours In the difference in the means sufficient to warrant an inference that type I is superior to type II regarding the length of life ? (Given /0.<)5,15 = 2.131)
1 2 6 | E n @m beh in g M athematics-111
16. The standard deviations calculated to n two random samples of size 9 and 13 are 2 and 1.9 respectively. May the sample be regarded as drawn from the nonnal populations with the same standard deviation. (Given Aw(s. «> * 3.51) 17. TWo sample are drawn from two normal populations. From die followong data test whether the two samples have the same variance at 5% level: Sample I 60 «S 71 74 76 82 85 87 Sample II 61 66 67 85 78 63 85 86 88 91 (Given f 9.9s (9. 7) * 3.68) 18. Following results were obtained from two samples, each drawn from two different populations A and B . Population A B Sample I II Sample Size n\ «* 25 rt2 = 17 Sampled s.
Opinion on autonomous status and level of graduation are independent Computed %2 = 1.622, and Drug is not better than sugar pills x2- 0-^8, Skilled fathers have intelligent boys. All the four dice are fair. (%2 * 1.845)
5.
6. 7.
x2 ~ 2.434, Drug is not effective. x2 ~ 16.666, The nature of work does not seem to be independent of the sex of the worker. The level and type of schooling are associated.
7. 9.
x2 = 6.77 tA= 1.66;/a = 0.85
S ta ti s tic s | 5 2 7
10.
/ = 0.85, No difference.
11.
t - 4; the sample mean differs significantly.
12.
t = 1.464, there is no reason to belive that the machine is defective.
13.
t - 1.937; the mean height of the students can be taken as 110 cm.
14.
/ = - 3.43 i.e., | f | = 3.43; No
15. / = 9.35, Type I bulb is superior to type II. 16. F ~ 1.15; the samples can be regarded as drawn from the population with the same standard deviation. 17.
F= 1.468, the samples have the same variance.
18. F - 2.205; variance of brand A is not more than the variance of brand B. 19. F = 1.415, the two populations have the same variance. 20.
z
= 0.086. □
5 2 6 | E ng in eerin g M a ih e m a iic s -III
TABLE AREA OF A STAMJAMBNORMAL DISTRIBUTION As entry in At total*Is Dmpmpmrtkm m&r «bt /I ottav am i wfelckii brtwn t ■ I ml t piMhc / I valmoft Aw wfcwrftiw nMnined / I bytym etry. z
.00
.01
... — .03 .02
0.0 0.1 02 03 0.4
.0000 .0398 .0793 .1179 .1554
,0040 .0438 .8832 .1217 1591
.0000 .0478 .0871 .1255 .1628
0.5 06 0.7 0.8 0.9
.1915 .2257 .2580 .2881 .3159
.1950 .2291 .2611 .2910 .3186
1.0 1.1 1.2 1.3 1.4
.3413 .3643 .3849 .4032 .4192
1.5 1.6 1.7 1.8 1.9
—^ -----£ .04
.05
.06
.07
.08
.09
.0120 .0517 .0910 .1293 .1664
,0160 .0557 .0948 .1331 .1700
J0199 .0596 .0987 .1368 .1736
.0239 .0636 .1026 .1406 .1772
.0279 .0675 .1064 .1443 .1808
.0319 .0714 .1103 .1480 .1844
.0359 .0753 .1141 .1517 .1879
.1985 .2324 .2642 .2939 .3212
.2019 .2357 .2673 .2967 .3238
.2054 .2389 .2703 .2995 .3264
.2088 .2422 .2734 .3023 .3289
.2123 .2454 .2764 .305! .3315
.2157 .2486 .2794 .3078 .3340
.2190 .2517 .2823 .3106 .3365
.2224 .2549 .2852 .3133 .3389
.3438 .3665 .3869 .4049 .4207
.3461 .3686 .3888 .4066 .4222
.3485 .3708 .3907 .4082 .4236
.3508 .3729 .3925 .4099 .4251
.3531 .3749 .3944 .4115 .4265
.3554 .3770 .3962 .4131 .4279
.3577 .3790 .3980 .4147 .4292
.3599 .3810 .3997 .4162 .4306
.3621 .3830 .4015 .4177 .4319
.4332 .4452 .4554 ..4641 .4713
.4345 .4463 .4564 .4649 .4719
.4357 .4474 .4573 .4656 .4726
.4370 .4484 .4582 .4664 .4732
.4382 .4495 .459! .4671 .4738
.4394 .4505 .4599 .4678 .4744
.4406 .4415 .4608 .4686 .4750
.4418 .4525 .4616 .4693 .4756
.4429 .4535 .4625 .4699 .4761
.4441 .4545 .4633 .4706 .4767
2.0 2.1 2.2 23 2.4
.4772 482! .4861 .4893 .4918
.4778 .4826 .4864 .4896 .4920
.4783 .4830 .4868 .4898 .4922
.4788 .4834 .4871 .4901 4925
.4793 .4838 .4875 .4904 .4927
.4798 .4842 .4878 .4906 .4929
.4803 .4846 .4841 .4909 .4931
.4808 .4850 .4884 .4911 .4932
.4812 .4854 .4887 .4913 .4934
.4817 .4857 .4890 .4916 .4936
Z5 2.6 2.7 Z8 19
.4938 .4953 .4965 .4974 .4981
.4940 .4955 .4966 .4975 .4982
.4941 .4956 .4967 .4976 .4982
.4943 .4957 .4968 .4977 .4983
.4945 .4959 .4969 .4977 .4984
.4946 .4960 .4970 .4978 .4984
.4948 .4961 .497! .4979 .4985
.4949 .4962 4972 .4979 .4985
4951 4963 4973 .4980 .4986
4952 .4964 .4974 .4981 498fc
3.0
.4987
\
4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 >4990
E xam . P a per s | i
a e . (Third Stouter) EnMinttM, Ji m , 2008 eacft UniftfrooMlpidsory. AH questions ifnifrl is aiMmatytic function and find its derivative. Ans. (Refer Page 22) f(°)
=
Aw. (Refer Page 59) C
Or (a) Evaluate the following integral using Cauchy’s integral formula; 4-3z dz z(z l)(z 2) c where C is the circle jzj = 3/2.
I
(b) Prove that:
l*2x
sio 2 0
2 ll
i «
. #j
= p r(a-V j -b ) where 0 < 6 < a.
Ans. (Refer Page 71) Ans. (Refer 93)
A2 , . A2sin(x+ h) ~£ + )+ Esin(x+ h) Ans. (Refer Page 202for a Similar Pmtetm) (b) Use Bessel’s formula to find the value ofy if x - 3.75, given: (a) Evaluate:
x y 25 24.145 3.0 22.043 33 20.225’ 4.0 11644 45 17.262 5.0 * 16J047; Ans. (Refer Page 239for a Similar Problem) Or (a) Find the first, second and third derivatives of the function tabulated below, at the pintx= 1.5: X is 2D 25 3.0 35 4.0
m 3375 7-0 13.625 24.0 38*75 59.0
Am. (Refer Page 260) (b) A body is in the form of a solid revolution. The diameter D in cm of the sections at distances x cm from one end are given below. Estimate the volume of the solid:
%|
E N G J^& O N G M A neM /m cs-U l
X..-. . D i t * . f i * i V/fi: *»*»*$,' v ~ * H* .v . • .-i< bni s VJ im SjO 6.0 s 'l ■’ ■ « .%« V. v •W *“ ■■ 8.75*' sir-1..- ‘ C * 10.0 625 55 12.5 15.0 4.0 nm> r *w jvt maiiwnw rrwmiy (a) By using Newton-Ramphson method, find the root o f x ^ x - 9 * 0 , which is near tox~ 2 correct to three places of decimal. Aas. (Rrfer P*$t 143) 2Jx+ 6 y - z m 85 6x + 15y + 2r - 72 x'+< 0 )= i AM.(g^er (b) Using Runge-Kutta method o f fourth order, solve
dy dx
v.
50/ jbr a Similar Problem)
y^ —X* ---- r with v(0)" 1 atx “ 0.2, 0.4. y*+x Am. (Xtfer Page 329)
Unit-IV
4.
(a) Find the rank o f:
' 6 4 10 16
1 2 3 4
3 8 % -1 9 7 12 15
(b) Show that the set S ® {(1,2, IX (2,1, OX (1, - 1,2)} forms basis of V^JX). Or (a) Show that the mapping/: V3 (K) -> defined as below /(a , A, c) ■ (c,
Exam. P apeib | 3
5.
[_• Unit-V (a) Prove that every Hermitian matrix # ean be represented in the form P+IQ, where P and Q are real symmetric and skew-symmetric metric? and phow th at: H * H is real if PQ “ - QP, t* 1 -1 2 -1 (b) Using Cayley-Hamilton theorem find the inverse 6f the matrix: A 1 -1 2 Or 3 -I I - 1 5 -1 to diagonal form. Hence (a) Find the matrix P which transforms the matrix A 1 -1 3 calculate if1. (b) Reduce foeanonical form and find the rank imd the signatwe o f the quadratic form : q ■ 1x2 -2 > ,2-3 * 2 - 4 yz+ 6zx
B.E. (Third Semester) EiM riM tjuit D c^ 20N (N) Mathematics-Ill . Note : The question paper ,is divided into five tmits. Each unit carries an internal choice. At tempt one question fiom each unit All questions carry equal marks. Uaift'I
I.
(a) Find the imaginary part oftheanalytic function/OO whose real part is x?-Ixy* +3x2 -3 y 2. Ans. (R tfer Page 2 1 ) *2+/ (b) Evaluate J ^ (2 jc+ + 1)<* along the two paths: . CO * ■ /+ U y - 2 f i - 1.
2.
(ii) The straight line joining ( 1 - 0 and (2 + /). A ns. (R efer Pmge 56 ) Or
1 1 (a) Find the image of the infinite strip ^
3.
Ans. (Rtfer Page 42)
f2*— ^ — * ■r ?mmmta > b > 0 Ans. (Refer Page 81) fo r a sim ilarproblem . Jo a + b cos0 Va2 -&2 Uatt-II (a) Use Newton’s divided difference formula to find a polynomial hi jc, given: jc 0 2 3 6 f(x) 648 704 729 792 Am*. (R efer Page 251) (b) Find the approximate value o f log. 5 by caicu&ing to four decimal places using Simpson's l/3rd rule for
f* dx
dividing the range into 10 equal parts. Or
Aas. (R efer Page 276)
4 | ENG1NEEHB4G MATHEMATfCS-lII
4.
(a) (i) Prove that:
( #
H
?
A ns, (R efer Page 204)
:
00 Estimate the missing termin foe following table: x: 0 1 2 3 4 /(x ). 1 3 9 ■ 81, Asa (Refer Page 213) (b) Find the first and second derivatives of the ahead function tabulated at the point x * 1 : x: ■ 1 3 5 7 9 f ( x ) : 85.3 74.5 67.0 60.5 54.3 Ans. (Refer Page 260) for a simalar problem. Unit-Ill 5.
(a) Use Newton-Raphson method to find a root Of die equation co sx -x 2 = x correct to four decimal places. Aas. (Refer Page 141)for a similar problem. (b) Use Runge-Kutta method to approximate the value ofy when x"m0.1 given that y (0) = and gfg 3*+ y
Or ^
Ans. (Refer Page 322)
(a) Solve the following system of equations by Jocobi’s iterative method correct to three decimal places: 10x-5>>-2z = 3 Ax - IO y + 3z = -3 x+6.y + IG* = - 3 Ans. (Refer Page 173) (b) Using Euler's method solve the differential equation for y at x = 1 in five steps: dy “£=jc2 + y2, j<0) = 1 Ans. (Refer Page 312)for a similar problem. 7.
Untt-IV (a) Apply Simplex method for solving the folk) ving L.P.P.: Maximize: z = 7x, + 5x2 Subject to :
x, + 2x2 56 4jc, + 3x 2 5 1 2
x ,,x2,£ 0 (b) Solve the following transportation problem: 01 02 Oi . Demand 8.
A 5 10 7 8
ih 3 7 5 12
JH 6 12 8 13
D< 5 4 -4 6
Supply 15 11 13
(a) Hie following table shows the efficiency of each operator handling different machines in terms of profit:
J E xam . P apeh » I 5
-* Machines Operators 4 I M hi
IV
9.
Mi
m2
M i-
m4
47 38 32 46
32 48 42 40
46 40 36 42
44 38 30 1 46
(b) Find the dua] of the following L.P.P: Minimize: z * x, + x2 + x3 Subject to; xt - 3x2 + 4x3 = i Xj - 2x2 £ 3 2x2-x3i 4 Xj.Xj ^O x3 is unrestricted in sign. »»it-V (a) Obtain the steady stale equations for the queueing model (M/M/1): (oo/FCFS). (b) Find the transition matrix corresponding to the following transition diagram:
Or (a) One type of aircraft is fqund to develop engine trouble in 5 flights out of a total of 100 and another type in 7 flights out of a total 200 flights. Is there a significant difference in the two types of aircrafts so far as engine defects are concerned ? (b) Write short notes on the following: (i) Sampling (li) Randomsmapling (iii) Factorial design (iv) Tfegpchi loss function B.E. (Third Semester) Examination, Jine, 2009 Matkematics-ni Note: Attempt all questions. All questions carry equal marks. One complete question solve at one place. 1. (a) Show that the Polar form of Cauchy-Reimann equations are : du 1 dv dv 1 du 10.
61 EwawcMBwam nK M tm xm ?r ' toVtfrr
+A~ -du ~+ $0? frl 58$
«■'' ‘.'-.-ir- • 1i A‘J
421 +z + 5 ^ ----------- - (
co f(35>
■;
m
Abs. (ReferPage 63) Or
(a) Expand:
m
1 (* - !)(*^2) ^ * e ,^ on W> 2.
A it (Refer Page 101)
cos26d6 2jw 2 f 2*_________________ j-(oJ < 1) Am. (Refer Page 89) *> l-2acos0+ a2 I (a) Show that the nth difference ofa polynomial of degrees will be constant and all (»+ l)thand higher ofder difference arc low. A*%. (Refer Page 200) lx= 1.1:
(b) Show that:
2.
X .
fix)
1J5 "'
6 0.128 0544 L296 2432 4jOOO
' 12
U 1j6
U 2J0
Ans. (Refer Page 266)
Or (a) Find;
3.
r
e*dx
by taking seven ordinates using Simpsoffe 1/3 mle. Ana. (Refer Page 285) (b) Define forward, backward, central difference operator and shift average operator also. (a) Solve the following by Euler modified method: ^ = log(x+y), y(0) * 2 at x » 1.4 with h « 0.2. A n. (Refer Page 317) (b) Find the roots of equation xe* = eosx using Regula-Falsi method, correct to four decimal places. Or Abs. (Refer Page 131) (a) Solve by Gauss-Seidel method: lOx, -2x2 -* j -x 4 * 3 -2xj + I0x2 - x 3~x4 «15 -x, - x 2+ 10*3 -24x4 » 27 -x, - x 2 -2xj + 10x4 m-9
Ana. (R efer Page 182)
V- v
B cmi. P apgbs { 7
(b) Giventhe values of u (x, y ) on the boundary'of the square in bloow. Evaluate foe function ‘ ‘‘i (x,$ safl&fymg foe Laplace equation ^ *0 aMhe Pivotal poWofthis figure. i© i
1000
1000. 1000
aooo
500
aooo
behB the lower
combined). Tbe belt A requires* fimcy buckle and only 400 per day are available. There are 700 buddes a day available for belt B. Deferable foe optimal product mix by graphical method, (b) Solve the following transportation problem: Destination Di Supply Da D* Oi 19 7 50 10 30 Sy Source 30 40 70 9 60 S2 70 20 40 8 18 S, Demand
5
8
14
7 Or
(a) Fourjobs are to be done on four different machines The cost (in rupees) o f producing fth job on the Jth machine is given below:
Machine Mx M2 M3 M4 A Job Ji A Ja
15 17 14 16
U 12 15 13
13 12 10 II
15 13 14 17
Assign the jobs to different machine so as to minimize the total cost
| E n g in eerin g M m h e m a iic s -ID
(b) Solve by Simplex method: Minimize: x f -3 u c j + 2 x 3
Subject to:
*rti .
3jcj ~ x 2 + 3x3 5 7 -2 x j + 4x2 512 ~4X| •+3x2 + 510 X|, Xj, x3 £0
(a) In a railway marshalling yard goods train arrive ata rate of 30 tram per day. If the distribution of arrivals is the Poisson and that of service time is exponential with an average 36 minutes,1 then find: (i) Mean queue size. (ii) The probability the queue size exceeds. If the input of trains increases to an average 33 per day what will be the change in (i) and (ii) ? (b) The state transition matrix’for retentions gains and -losses of firms A, B and C is given below. Using this matrix determine the steady state equilibrium condition: . Form To A B C A 0.700 0.100 0.200 B 0.100 0.800 0.100 C 0.200 o.ioo 0.700 ! Or (a) Write the advantage and disadvantage of Robust design method. (b) The number of units of an item that are withdrawn from inventory on a day to day basis is a Markov Chain Process in which requirement for tomorrow depend on today’s requirement A one day transition matrix is jpven below. Number of units withdrawn from inventory : Tomorrow 5 10 12 5 0.6 0.4 0.0 Today 10 03 03 0.4 12 0.1 03 0.6 (i) Construct a tree diagram showing inventory requirements on two consecutive days. (ii) Develop a two day transition matrix.
E xam. P a fb k } 9
B.E* (Third Semester) Fiawlasrton, Feb, 2010 Matfceraatks-IU Note : Attempt five questions in all selecting one question from each unit All questions cany equal marks.
U nft-I J.
(a) Show that the function e* (x cos y - y siny) is harmonic and find its conjugate. Abs. (Refer Rage 24) 2:
-
•
where C is the circle Iz 1*3. (b) Evaluate f ----*c ( z - lX * -2 ) 2.
Abs. (Refkr Page 76)
Or (a) Find the bilinear transformation which maps the joints * = l , i , - 1 onto points w * 4 0, - i. 1 Abs. (Refer Page 25) r2/% COS20dB (b) Evaluate ! ~—r----- r— r* (tf2< t.) Jo l-2tfcos6+
Abs.
Page 89)
U nfc-II 3.
(a) (i) Prove that:
Abs. (Refer Page 204) (ii) Obtain the function whose first difference is 2X3 + 3x* - 5x + 4.Abs. (Refer Page 211) dy (b) Find ~ at x * 1.1 from the following table:
X
’ TF 12 1.4 1.6 1* 2.0
4.
y 0
0.128 0544 1296 2.432 ' 4.00
Or (a) Find the cubic polynomial which takes the following values x : 0 1 . . 2 3 f(x) : 1 2 1 10 Hence or otherwise evaluate/(4).
Abs. (Refer Page 266)
A bs. (R efer Page 223)
1 0 1EmMHOHQKwMAncs-ill (b) The following tabfe gives the vetority vofapartjcleattinK;;;
-
r
2 4 6 8 10 12
! v (m W .i v " ■' . *■' ' *6 16 34 60 ' 94 136
Find the distance moved by the particle in 12 seconds and also the acceleration at/-»2 second. Ass. (Refitr Page 28$)
Unit-III 5
6.
(a) Find a real root of the equation jc log* x ~ 1.2, correct to five decimal places by NewtonRaphson method. Ans. (Refer Page 146) (b)Sotvetheequdiom: Sx+2 y + z * 12 x + 4 y + 2 z * 15 x + 2 ^ + 5 z * 20 by Qmm Saidel method. Ans. (Refer Page 185) Or (a) Solve the equations: 3x + y + z » 4 x+2y+2z * 3 * • 2x+y+3z * 4 byCroutVmethod (b) Apply Runge-Kutta method to find an approximate value of >»for x * 0.2 in steps of 0.1, if *dy — “ X+^.^CO)* 1.
Ans.(ReferPage321) U r it-IV
7.
(a) Solve the following L.P.P., using Simplex method: Maximize z*5xj + 3xj Subject to : x\ +x2 £ 2 5xi + 2x2 5 10 3xi + 8x2 5 12 xi,x2 it 0
11
*’*L
(b) Solve the foi O m s u i ^ ^ ..A” ‘ I 11 ni
6 4 I
a 9 2
4 8 6
14 12 5
Required
6
10
15
31
,
8
Available
* ir
i
*' v
(a) Using duality solve die following problem: Maximize: •z **0-7 x, + 0.5 x 2 Subject to : x i ^ 4 , x2 ^ 6 . xj + 2xi * 20 2&1+X2 £ 1 8
.
*>' >■ ■
.
X |,X 2 * 0
(b) A firm plans to begin production of three new products on its. three ptftst*. H eiptt cost of producing / at plantj is as given below. Find the assignment that Minimises the total unit cost: Plant ’ Product
1 2 3
1 10 18 6
2 8 6 4
* 3 12 14 2
U n it -V
9.
10.
(a) Obtain the steady state equations for the (M j M11) : (oo /FCFS) qticuing model. (b) Write sfaoit notes on the following: (i) Taguchi toss function (K) Markovian process Or (a) There is congestion on the platform of railway station. The trains arrives at the rate of 30 trains per day. The waiting time for any train to hump is exponentially distributed with an average of 36 minutes. Calculate the mean queue size and the probability that queue size exceeds 9. (b) Write short notes on the following: 0) Factorial design (ii) Robust design method.
1 2 | E n q b « e8 b « 3 M a th e m a tic s-!!!
B.E. (Third Semester) E uniaitkm , J u e , 201! Mathemattes-III Note: Attempt all questions by selecting any two part*\fi*« e*ch Questions. All questions carry equal marks. (a) Find the imaginaiy part of the analytic function who^e/eal part is : 1. Ans. (Refer Page 21) x 3 -3 ^ -3 * a -3 ^ (b) Evaluate: f _ £2z... dz k < * + l)4 Ans. (X ^ r
where C is the circle 12 j * 3. (c) Show that: r*«__£ *0 o 2 + cos0 (d) Show that the transformation co “
.
2
2it ^
'
6/)
Ans. (Refer Page 81)
2*+3 ^ changes the circle x1 +>*-4z * 0 into straight line Ans. (ffgfer
4* + 3 ~ 0, (a) Prove that: (i) e * > » l + A
5??
Ans. (Refer Page 198) Ans. (Rtfer Page 204)
09 e* (b) Find/(9 ) from the following table: x 5 7 11 13 17
/<*) 150 392 1452 2366 5202
Ans. (Rtfer Page
>
(c) Find — atx * 1.5 from the following table: x 1.5 2.0 2.5 3.0 15 4.0
y 3.375 7.0 13.625 24.0 38.875 59.0
A bs. (R efer Page 260)
• A EXAK. PAFtMfi | IS (d) Evaluate:
"V m/2 .
^ *'* *’*' -*" viV/
fU
: . . t,
... \ . .
,.*■
(i) Rising Simpsons - ~ rule.
i: . i
u,
, • • . .
(ii) Using Weddle’s rule. Aim*.(Refer Page 277) (a) Using Newton-Raphson method find the real root of the equation 3x * cos x + 1 correct to five decimal places. Abs. (Refer Page 141) (b) Solve the following system of Aquations using Gauss-Seidel method : 27x + 6 y -* * 85 6x + \Sy + 2z “ 72 x + y + 54z « 110 Ans.(ReferPage 179) dy ‘ . (c) Findy (2.2) using Euler’s method from the equation xy*9y (2) * 1. Ass. (Refer Page 313) (d) Apply Runge-Kutta method to find an approximate value ofy for x » 0.2 in steps of 0.1 if dy
-r£ pcx+^a, given thaty* 1 whenx~0.
Ans. (Refer Page 321)
(a) Solve the following L.P.P. by Simplex method: Max: z - 3r|+2r2 Subject to: X| + Xj £ 4
x j-x j £i 2 X |,X 2 fc'O
(b) Solve the following L.P.P. by Graphical method: Max.: z * 5xi + 7xj Subject to constraints: xj + x2 S 4 3xt + 8x2 £ 24 10bC| + 7x2 £ 35 xn x2 * 0 (c) Find the initial basic feasible solution of the following transportation problem by Vogel’S approximation method: Wvehouse
Factory
W, F, 19 F2 70 F, * 40. 5
w, 30 30 8 8
w, 50 40 70 7
w4 10 60 20 14
7 9 18 34
minutes. If be mpaks scts in oif4fryta which ti&y come maad if ike arrival of sets Is approxi mately Potoon within averagfrfjats of 10 per 8 hours perday, what is repairman's expected idfe time each day? How manyjobs are ahead of the average set just brought in j — (c) Write short note on the foUowbgg: N^ toetioii R o^dw gnoK ttxxL ?f, U t (Third Stme&er) fcuwhurtlee, Dee, 2011 ii* fiom each Unit All question carry equal marks! (Jlitt-I (a) Shew that the frootiong^x3-la y 1 is harmonic andlind the corresponding analytic function afthisas the*Wpait ( Ass. (Refer Page 20) fh) jjl^ lTed flflrivn ^airTij theorem. Aaa. (3Iq^tr A ge 5*> Or (a) Find the b l t & i e a ^ j s u q a s U iv*-1 to /, 0, - / respectively. > 1 Abs. (Ifgfcr Page 35) *
A
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f oosxr « integral formula, evaluate the integral1 L 7' ’n ;r .’n where Cis the circle .*; • . • • • ' . ; ’■.* . -iy ;vv’■' U"■ ‘VV Ans. (K g tr/to ? * 77) UttU-H (a) Prove with usual notations that: AD " log (1 + A) * - tog (1 - V) (b) Expressy-2X3 - 3X2 + 3Lc- 10 in fectoriri notation form. Or (a)G|ventiiat: x y . 1.0 7.989 1.1 8.403 8.?8l U ■/* 1.3 9.129 9.451 1.4 F.5 9.750 1.6 10.031
Find ffe dx
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A bs. (R tfe r Page 242)
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Am * Or (a) Solve thesystem ofequations by Qqm*G*toi ftyeffooafeod: . i . ., 27* + 6 > - z » | 5n-2z - 72 .?' > + * + 54* « 110, ,, . ^ (b) By using NeftftteRaphson mated, find* rpatf** of*ie S ^ p t t a i i ^ w f e f e f c li. n^ar to * » 2#Mitect to three deotal places, " IMIttMdW W M V (a) Find th v M v f ftr<owing L.P.P.: Mf*%jr+zx - 3 y + 4z *>.5 '• .. x + 2y 5 . 3 2x-ii4 x,^,z £ 0 (b) Four jobs are to be demon four (Sfiefeat xaacktnes. The cast 0n rupees) oCfKofecmg Alt job toyth machine is gh*a below: V m kim — ►
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(b) Solve the following L.P.P. by Graphical method : Maximize : z = 3x + 2y Subject to: 2x+y £ 5 x+y £ 3 x,y * 0
U iiit-V (a) Write advantage and disadvantage of Robust design method (b) in a railway marshalling yard goods train arrive at a rate of 30 trains per day. If the distribute of arrivals is Poisson and that of service time is exponential with an average 36 minutes, the find: (i) Mean queue size (ii) The probability that queue size exceeds. Or (a) Describe the differentitiMifference equations for the queuing model (M/M/1) : (oo/qo/FCFS (b) Write short notes on the following: (i) Taguchi loss function (ii) Factorial design.
The subject Engineering Mathematics is the backbone of a degree in any field of engineering and occupies a prom inent place in the schemes and syllabi of technical institutes and universities. While fram ing the content and m atter of the book special attention has been paid to the em erging demand of engineering students in the field of mathematics. The book in its present form covers topics and examples,, both solved and unsolved exercises, useful for students’ practice and explanatory purposes. KEY FEATURES OF THE BO O K
An atte m p t has been made to assure th a t the book caters the need of engineering students. Worked examples and graded exercises are used throughout to develop ideas and concepts. The form at of this book makes it an easy companion fo r the basis fo r a complete m odular course in Engineering Mathematics. The book covers following topics for the second year engineering student
• • • • • •
Functions of Complex Variables Numerical Analysis Interpolation Correlation and Regression Theoretical Distribution Testing of Hypothesis
Dr. D.K. Jain did his M.Sc. in Mathematics in the year 1991 from Jiwaji University, Gwalior. He has obtained M.Phil. Degree in the year 1992. He has also cleared the UGC JRF-NET Exam for Lecturership and was awarded Junior Research Fellowship for his Ph.D. degree from Jiwaji University in the year 1997 on Special Function and its Applications. He has published several research papers in national and international Journals and anthologies besides having authored a full series of books for the Engineering students on engineering mathematics, useful for the Undergraduate as well Post-graduate students. He has a Post-graduate teaching experience of over Sixteen years and is presently teaching in the Department of Applied Mathematics at Madhav Institute of Technology and Science, Gwalior, a renowned technical institute of Madhya Pradesh.