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Question 1 a) Each row corresponds corresponds to a diferent observation, i.e. i.e. each row represents represents the data o each o the 100 employees. There is 1 continuous variable (!). "onthly salary salary in dollars is a measured amount. There are various units o measurement or money (dollars and cents) and also, observations or salary are listed to two decimal points. Thereore Thereore it is a continuous variable. There are # nominal variables (1, $, %).There are
b) The histo&ra histo&ram m is unimodal unimodal with a sli&ht sli&ht positive positive s'ew s'ew.. Histogram of Monthly Monthly Salar Salary y 1% 1( 1! y 10 c n e u ) q e r F
% ( ! 0
!$00
#000
#$00
(000 ($00 Monthly Salary
c) *emales +ower median -li&h -li&htl tly y lar lar&e &err int inter eru uar arti tile le ran& ran&e e -li&htly ne&atively s'ewed o outliers -maller ran&e +ower maimum value +ower minimum value
$000
$$00
"ales i&her median -li& -li&ht htly ly smal smalle lerr int inter eru uar arti tile le ran& ran&e e /ositively s'ewed Three outliers +ar&er ran&e i&her maimum value i&her minimum value
d)
The &raph shows a linear relationship with a positive correlation. orrelation can be epected to be positive and lie between 0.$ and 1, closer to 0.$. e) ovariance between monthly salary and len&th o service is 1%$.$.
covariance = Correlation ) the product of standard deviations ρ=
¿
cov ( x , y ) σ x σ y
1654.5
√ 445327.05 × √ 17.26
¿ 0.59677
(to $ si&. 2&.)
3 correlation o approimately 0.% is con2rmed by the scatterplot rom (d). 4t shows that there is a noticeable positive linear relationship between monthly salary and len&th o service. &)
Scatterplot of Monthly Salary vs Length of Service %000
5ender *emale "ale
$000 y r a l a S y (000 l h t n o M
#000
!000 $
10
1$ !0 Length of Service
!$
#0
The claim is true. *rom this scatterplot it can be clearly seen that both &enders have a more distinct and more pronounced positive linear relationship. The linear trend or males is parallel and consistently hi&her than the linear trend or emales.
h)
Empirical CDF of Monthly Salary 100
)0
t %0 n e c r e P (0
!0
0 !000
#000
(000 Monthly Salary
$000
%000
i. !6.17 (to 1 d.p.) o employees have a monthly salary less than 8#$00. ii. $7 o employees have a salary over 86%$.!0 (to the nearest 10 cents).
Question ! a) Table o wor'in&9 "ale 0.$$?0.$ 0.%A0.$$?0.$
lerical Technical
0.%
*emale (1A0.$$)?0.$ 0.$A(0.%A 0.$$?0.$) 0.
0.$ 0.$ 1
Table o probabilities9 "ale 0.!6$ 0.#!$ 0.%
lerical Technical
*emale 0.!!$ 0.16$ 0.
0.$ 0.$ 1
Thereore, the probability that an employee is a clerical staf member and is male is 0.!6$. b) /(clerical staf:male) ; /(clerical < male) = /(male) ; 0.!6$=0.% ; 0.$# (to d.p.) Thereore, the probability that the employee is a clerical staf member &iven that they are 'nown to be male is 0.$#. c) 4ndependence test9 /(3∩>) ; /(3) ? /(>) /(clerical < male) ; 0.!6$ /(clerical) ? /(male) ; 0.$ ? 0.% ; 0.# /(clerical < male)
≠
/(clerical) ? /(male)
Thereore, &ender and @ob classi2cation are not independent.
d)
i. There were $# clerical staf members included. ii. There were $ male staf members included. iii. There were #! male, clerical staf members included. e) B would be a nominal random variable. -ince B is the variable o an individual, it has only one trial, &ivin& it a >ernoulli distribution (and not a binomial distribution). f) MTB > Set C1 DATA> 1( 1 : 100 / 1 )1 DATA> End.
C is a discrete count variable which can ta'e inte&er values rom 0 to 100. CD>in(n;100, p;0.$). h) i) E(C) ; np ; 100?0.$ ; $0 Epected value is $0. ; Fnp(1Ap) ; F100?0.$?(0.$) ;$ -tandard deviation is $. $) MTB > De%&r"'e C2; SUBC> Sm%.
Descriptive Statistics: C2 ar"a'!e C2
Sm *.0000
The observed value o C (the total number o the &roup who are classi2ed as technical) is # (since technical @obs were coded as 1 and clerical @obs were coded as 0). ') i. +et the total number o the sample &roup who are classi2ed as clerical be C. Gbserved value o C rom spread sheet values is $6. /(CH$6) ; /(CI$%) MTB > CD+ #,; SUBC> B"nom"a! 100 0.#.
Cumulative Distribution Function B"nom"a! -"t n 100 and 0.# #,
( 3 4 ) 0.50**2, Terefore6 ro'a'"!"t7 of tat tere -o!d a8e 'een !e%% tan #9 &!er"&a! %taff "n a %am!e of 100 "% 0.50**2,.
ii. +et the total number o the sample &roup who are classi2ed as clerical be C. Gbserved value o C rom spread sheet values is $6. /(CJ$6) ; 1 A /(CI$6)
MTB > CD+ #9;
SUBC>
B"nom"a! 100 0.#.
Cumulative Distribution Function B"nom"a! -"t n 100 and 0.# #9
( 3 4 ) 0.5***5#
/(CI$6) ; 0.K###K$ /(CJ$6) ; 1 L 0.K###K$ ; 0.0%%%0$ iii. MTB > D+ #9; SUBC> B"nom"a! 100 0.#.
Probability Density Function B"nom"a! -"t n 100 and 0.# #9
( 3 ) 0.0*00,,
Terefore ro'a'"!"t7 of tere 'e"ng ea&t!7 #9 &!er"&a! %taff mem'er% "n a %am!e of 100 "% 0.0*00,,.
Question # i. M is a discrete random variable. M ! # $ % 6 K 10 11 1! Total
N ; !?1=#% O #?1=1 O ?1=1! O $?1=K O %?$=#% O P O 10?1=1! O11?1=1 O 1!?1=#% ; 6 Thereore the mean is 6.
∑ x P ( x )− μ 2
!
(C) ; ;
2
all x
(C) ; !?!?1=#% O #?#?1=1 O P O 11?11?1=1 O 1!?1!?1=#% L 6?6 ; 6.## ii. *irst roll ; $ -econd roll ; % Gbserved value o M ; 11 iii. /robability density unction9 *(") ; 1=0 where 0 I " I 0 Distribution Plot Rniform, +ower;(0, Rpper;)0 0.0!$
0.0!0
y 0.01$ t i s n e D
0.010
0.00$
0.000 (0
$0
%0 M
60
"ean ; E (") ; (aOb)=! ; (0O0)=! ; %0 (") ; (bAa)S! = 1! ; (0A0)S! = 1! ; 1##.## iv. "ean o amount received ; E(" ? M) ; M ? E(") ; 11 ? %0 ariance o amount received ; (M?") ; MS! ? (")