A recent census of employees at a large automotive company showed that 30% of th

Question

A recent census of employees at a large automotive company showed that 30% of the assembly line workers were extremely bored with their job. The management of the company arranged for a consulting firm to come and offer a program to reduce on-the -job boredom. After two months of the program, a random sample of 13 workers was selected to help evaluate its impact.

(a) If the program was not effective, what is the probability that exactly 6 workers in the sample would report being bored with their job? (First translate words into symbols – you have to do this for a quiz or exam)

b) If the program was not effective, what is the probability that more than 6 workers in the sample would report being bored with their job?

c) If the program was not effective, what is the probability that at most 6 workers in the sample would report being bored with their job?

d) If the program was not effective, what is the probability that more than 5 but at most 9 workers in the sample would report being bored with their job?

e) If the program was not effective, what is the probability that at least 5 but not more than 9 workers in the sample would report being bored with their job?

f) Suppose that that among 13 randomly workers, two of them would report being bored with their job. Is there any evidence to suggest that the program is effective? Justify your answer with calculations.

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 13 * 0.3
= 3.9
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 13 * 0.3 * 0.7
= 2.73
III.
standard deviation = sqrt( variance ) = sqrt(2.73)
=1.6523
a.
P( X = 6 ) = ( 13 6 ) * ( 0.3^6) * ( 1 - 0.3 )^7
= 0.103
b.
P( X < = 6) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 13 6 ) * 0.3^6 * ( 1- 0.3 ) ^7 + ( 13 5 ) * 0.3^5 * ( 1- 0.3 ) ^8 + ( 13 4 ) * 0.3^4 * ( 1- 0.3 ) ^9 + ( 13 3 ) * 0.3^3 * ( 1- 0.3 ) ^10 + ( 13 2 ) * 0.3^2 * ( 1- 0.3 ) ^11 + ( 13 1 ) * 0.3^1 * ( 1- 0.3 ) ^12 + ( 13 0 ) * 0.3^0 * ( 1- 0.3 ) ^13
= 0.9376
P( X > 6) = 1 - P ( X <=6) = 1 -0.9376 = 0.0624
c.
P( X < = 6) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 13 6 ) * 0.3^6 * ( 1- 0.3 ) ^7 + ( 13 5 ) * 0.3^5 * ( 1- 0.3 ) ^8 + ( 13 4 ) * 0.3^4 * ( 1- 0.3 ) ^9 + ( 13 3 ) * 0.3^3 * ( 1- 0.3 ) ^10 + ( 13 2 ) * 0.3^2 * ( 1- 0.3 ) ^11 + ( 13 1 ) * 0.3^1 * ( 1- 0.3 ) ^12 + ( 13 0 ) * 0.3^0 * ( 1- 0.3 ) ^13
= 0.9376
d.
P(5 < X < = 9) = P(X=9) + P(X=8) + P(X=7) + P(X=6)
= ( 13 9 ) * 0.3^9 * ( 1- 0.3 ) ^4 + ( 13 8 ) * 0.3^8 * ( 1- 0.3 ) ^5 + ( 13 7 ) * 0.3^7 * ( 1- 0.3 ) ^6 + ( 13 6 ) * 0.3^6 * ( 1- 0.3 ) ^7
P(X < X <= 9) = 0.103 + 0.0442 + 0.0142 + 0.0034 = 0.1648

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