Question

the army gives a battery of exams to all new recruits. one exam measures a person's ability to work with technical machinery. this exam was given to a random sample of 360 new recruits. in the table below x-scores on the test and f is the frequency of new recruits with score x.

x 1 2 3 4    5 6 7    8    9    10

f    28    42 79 83    51 36 18    12    7    4

a) find the probability distribution for these scores

b) draw a histogram of the probability distribution

c). the ground to air missile battalion needs people with a score of 7 or higher on the exam. What is the probability that a new recruit will meet this criterion?

d). the kitchen battalion can use people with scores of three or less. What is the probability that a new recruit is not overqualified for this work?

E). compute the expected value of these scores

f). compute the standard deviation of these scores

Answer 1

a) Let X denotes the scores on the test.

The probability distribution of X is

X	Frequency	P(X=x) = ( Frequency / Total frequency)	x* P(X=x)	x2* P(X=x)
1	28	0.0778	0.0778	0.0778
2	42	0.1167	0.2333	0.4667
3	79	0.2194	0.6583	1.9750
4	83	0.2306	0.9222	3.6889
5	51	0.1417	0.7083	3.5417
6	36	0.1000	0.6000	3.6000
7	18	0.0500	0.3500	2.4500
8	12	0.0333	0.2667	2.1333
9	7	0.0194	0.1750	1.5750
10	4	0.0111	0.1111	1.1111
Total	360	1.0000	4.1028	20.6194

b) Histogram

Using R

> x=1:10
> f=c(28,42,79,83,51,36,18,12,7,4)
> y=rep(x,f)
> b=seq(0.5,10.5,1)
> hist(y,breaks=b,xlab="X", ylab="frequency",main ="Histogram")

frequency 0 20 40 60 80 Histogram

c) Required probability = P ( X> =7)

= P ( X=7) + P( X=8) + P(X=9) + P(X=10)

= 0.0500 + 0.0333 + 0.0194 + 0.0111

= 0.1139

d) Required probability = P ( X < = 3)

= P ( X=1) + P(X=2) + P(X=3)

= 0.0778 + 0.1167 + 0.2194

= 0.4139

e) Expected value of scores

E(X) =sum ( x * P(X=x)) = 4.1028

f) Var(X) = E(X²) - (E(X)) ²

E(X²) = sum ( x² * P(X=x)) =20.6194

Var(X) = 20.6194 - 4.1028² = 3.7867

Standard deviation of the scores is

SD(X) = sqrt(Var(X)) = 1.9459.