Unit# 6 Basic Statistic Exercise 6.2

Unit 6 Basic Statistic Exercise 6.2 Question # 3 Find Arithmetic mean by direct method for the following set of data: (i)

12, 14, 17, 20, 24, 29, 35, 45.

Solution

= =

= = 24.5 (ii)

200, 225, 350, 375, 270, 320, 290.

Solution

= =

= = 290

Mudassar Nazar Notes Published by Asghar Ali

Page 1

Question # 4 For each of the data in Q# 3 Compute arithmetic mean using indirect method

(i)

12, 14, 17, 20, 24, 29, 35, 45.

Solution

A

X 12 14 17 20 24 29 35 45

D = x 20 -8 -6 -3 0 4 9 15 25 = 36 –

=A+ = 20 + = 20 + 4.5 = 24.5 (ii)

200, 225, 350, 375, 270, 320, 290.

Solution

A

X 200 225 270 290 320 350 375


D = x 290 -90 -65 -20 0 30 60 85 =0 –

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=A+ = 290 + = 290 + 0 = 290

Question # 5 The marks obtained by students of class XI in mathematics are given below compute arithmetic mean by direct and indirect methods. Classes/ Groups 0 9 10 19 20 29 30 39 40 49 50 59 60 69

Frequency 2 10 5 9 6 7 1

–

–

–

–

–

–

–

Solution

Classes/ Groups 0 9 10 19 20 29 30 39 40 49 50 59 60 69 –

–

–

–

–

–

–

f 2 10 5 9 6 7 1 = 40

X 4.5 14.5 24.5 34.5 44.5 54.5 64.5

fX 9 145 122.5 310.5 267 381.5 64.5 X= 1300

Direct Method =

= = 32.5 Mudassar Nazar Notes Published by Asghar Ali

Page 3

Class/ Group 0 9 10 19 20 29 30 39 40 49 50 59 60 69 –

–

–

–

–

–

–

f 2 10 5 9 6 7 1 = 40

X 4.5 14.5 24.5 34.5 44.5 54.5 64.5

D = x 34.5 -30 -20 -10 0 10 20 30 –

fD -60 -200 -50 0 60 140 30 = -80

Short cut Method ( Indirect Method) =A+ = 34.5 + = 34.5 - 2 = 32.5

Class/ Group 0 10 20 30 40 50 60

–

–

–

–

–

–

–

9 19 29 39 49 59 69

f

X

2 10 5 9 6 7 1 = 40

4.5 14.5 24.5 34.5 44.5 54.5 64.5

u= -30 -20 -10 0 10 20 30

fu -6 -20 -5 0 6 14 3 = -8

Coding Method ( Indirect Method) =A+

xh

= 34.5 + = 34.5

–

10 2

= 32.5


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Question # 6 The following data relates to the ages of children in a school. Compute the mean age by direct and short cut method taking taking any provisional provisional mean ( Hint take A = 8) Class limits 4 6 7 9 10 12 13 15 Total Also compute Geometric mean and Harmonic mean.

Frequency 10 20 13 7 50

–

–

–

–

Solution

Class limits 4 6 7 9 10 12 13 - 15

f 10 20 13 7 = 50

–

–

–

X 5 8 11 14

fX 50 160 143 98 X= 451

Direct Method =

= = 9.02

Class limits 4 6 7 9 10 12 13 - 15 –

–

–

F 10 20 13 7 = 50


X 5 8 11 14

D=x 8 -3 0 3 6 –

fD -30 -30 0 39 42 = 51

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Short cut Method ( Indirect Method) =A+ =8+ = 8 + 1.02 = 9.02

Class limits 4 6 7 9 10 12 13 - 15 –

–

–

F 10 20 13 7 = 50

X 5 8 11 14

logx 0.6990 0.9031 1.0414 1.1461

f logx 6.990 18.062 13.5382 8.0227 = 46.6129

0.2 0.125 0.0909 0.0714

2 2.5 1.1817 0.4998

G. M = Anti-log G.M = Anti-log G.M = Anti-log ( 0.9323) G.M = 8.56 Class limits 4 6 7 9 10 12 13 - 15 –

–

–

f

X

10 20 13 7 = 50

5 8 11 14

= 6.1815

H.M = H.M = H.M = 8.089


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Question # 7 The following data shows the number of children in various families. Find Mode and

Median. 9, 11, 4, 5, 6, 8, 4, 3, 7 ,8, 5, 5, 8, 3, 4, 9, 12, 8, 9, 10, 6, 7, 7, 11, 4, 4, 8, 4, 3, 2, 7, 9, 10, 9, 7, 6, 9, 5. Solution

Arranged Data

(n = 38) 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5 , 5, 5, 5, 6, 6, 6, 7, 7 , 7, 7, 7, 8, 8, 8, 8, 8 , 9, 9, 9, 9, 9, 9, 10, 10, 11, 11, 12. Mode = the most frequent number Mode = 4 , 9 Median = [

th

observation +

th

observation +

th

observation ] th

Median =

[

Median =

[ 19 observation + 20 observation ]

Median =

[ 7 + 7]

Median =

[14]

th

observation ]

th

Median = 7


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Question # 8 Find modal number of heads for the following distribution showing the number of heads

when 5 coins are tossed. Also determine Median. X ( Number of heads) 1 2 3 4 5

Frequency ( number of times) 3 8 5 3 1

Solution

X 1 2 3 4 5

f 3 8 5 3 1

C.F 3 11 16 19 20

= 20

n= Mode = the most frequent observation Mode = 2 Median Median = the class class conta containi ining ng ( )

th

Medi Median an = the the clas classs conta contain inin ing g ( )

observation th

observation

th

Median = the class containing 10 observation Median = 2

Question # 9


Page 8

The following frequency distribution is the weights of boys in kilograms. Compute mean, median and mode. Class Intervals 1 3 4 6 7 9 10 12 13 15 16 18 19 21

Frequency 2 3 5 4 6 2 1

–

–

–

–

–

–

–

Solution

Class Intervals 1 3 4 6 7 9 10 12 13 15 16 18 19 21 –

–

–

–

–

–

–

f 2 3 5 4 6 2 1 = 23

X 2 5 8 11 14 17 20

fX 4 15 40 44 84 34 30 =241

Mean =

= = 10.48


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Class Intervals 1 3 4 6 7 9 10 12 13 15 16 18 19 21 –

–

–

–

–

–

–

f 2 3 5 4 6 2 1 = 23

C.F 2 5 8 11 14 17 20

C. Boundaries 0.5 3.5 3.5 6.5 6.5 9.5 l 9.5 12.5 12.5 15.5 15.5 18.5 18.5 21.5 –

–

–

–

–

–

–

= = = 11.5 Median = l

+

[

- c]

Median = 9.5 + ( 11.5

–

10 )

Median = 9.5 + ( 1.5) Median = 9.5 + 1.125 Median = 10.625

Class Limits 1 3 4 6 7 9 10 12 13 15 16 18 19 21

f 2 3 5

–

–

–

–

–

–

4 6 2

–

–

–

f 1 f m f 2 1

–

C. Boundaries 0.5 3.5 3.5 6.5 6.5 9.5 9.5 12.5 12.5 15.5 15.5 18.5 18.5 21.5 –

–

–

–

= 23

Mode = l + Mudassar Nazar Notes Published by Asghar Ali

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Mode = 12.5 + Mode = 12.5 + Mode = 12.5 + Mode = 12.5 + 1 Mode = 13.5 Question # 10 A student obtained the following following marks at a certain examination. examination. English 73, Urdu 82, Maths 80, History 67 and Science 62. (i)

If the weights accorded these marks are 4, 3, 3, 2 and 2 respectively, what is an appropriate average marks?

(ii)

What is the average mark if equal weights are used?

Solution X (marks) 73 82 80 67 62 = 364

w ( weight) 4 3 3 2 2 = 14 (i)

Xw 292 246 240 134 124 w = 1036

Weighted Mean

w= w= w = 74 (ii)

Arithmetic Mean

= = = 72.8


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Question # 11 On a routine trip a family bought 21.3 liters of petrol at 39.90 rupees per liter, 18.7 liters at 42.90 rupees per liter and 23.5 liters at 40.90 rupees per liter. Find the mean price paid per liter.

Solution

Mean price = Mean price = Mean price = Mean price = 41.15 rupees per liter.

Question # 12 Calculate simple moving average of 3 years from the following data: Year Value

2001 102

2002 108

2003 130

2004 140

2005 158

2006 180

2007 196

2008 210

2009 220

2010 230

Solution

Year

Value

2001 2002 2003 2004 2005 2006 2007 2008 2009 2010

102 108 130 140 158 180 196 210 220 230


3 Year Moving Total

Average

…….

…….

340 378 428 478 534 586 626 660 -------

113.33 126 142.67 159.33 178 195.33 208.67 220 ------

Page 12

Unit# 6 Basic Statistic Exercise 6.2

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