In a survey conducted by the Society for Human Resource Management, 68% of worke

Question

In a survey conducted by the Society for Human Resource Management, 68% of workers said that employers have the right to monitor their cell phone use. Suppose that a random sample of 15 workers is selected, and they are asked if employers have the right to monitor their cell phone use. a). What is the expected number of workers to agree? What is the standard deviation? b). What is the probability that exactly 5 workers agree? c). What is the probability that more than 10 workers agree? d). What is the probability that less than 5 workers don’t agree?

Explanation / Answer

Binomial 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
a.
Mean ( np ) =15 * 0.68 = 10.2
b.
Standard Deviation ( npq )= 15*0.68*0.32 = 1.8067
c.
P( X = 5 ) = ( 15 5 ) * ( 0.68^5) * ( 1 - 0.68 )^10
= 0.0049
d.
P( X < = 10) = P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 15 10 ) * 0.68^10 * ( 1- 0.68 ) ^5 + ( 15 9 ) * 0.68^9 * ( 1- 0.68 ) ^6 + ( 15 8 ) * 0.68^8 * ( 1- 0.68 ) ^7 + ( 15 7 ) * 0.68^7 * ( 1- 0.68 ) ^8 + ( 15 6 ) * 0.68^6 * ( 1- 0.68 ) ^9 + ( 15 5 ) * 0.68^5 * ( 1- 0.68 ) ^10 + ( 15 4 ) * 0.68^4 * ( 1- 0.68 ) ^11 + ( 15 3 ) * 0.68^3 * ( 1- 0.68 ) ^12 + ( 15 2 ) * 0.68^2 * ( 1- 0.68 ) ^13 + ( 15 1 ) * 0.68^1 * ( 1- 0.68 ) ^14 + ( 15 0 ) * 0.68^0 * ( 1- 0.68 ) ^15
= 0.5523
P( X > 10) = 1 - P ( X <=10) = 1 -0.5523 = 0.4477
e.
P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 15 4 ) * 0.32^4 * ( 1- 0.32 ) ^11 + ( 15 3 ) * 0.32^3 * ( 1- 0.32 ) ^12 + ( 15 2 ) * 0.32^2 * ( 1- 0.32 ) ^13 + ( 15 1 ) * 0.32^1 * ( 1- 0.32 ) ^14 + ( 15 0 ) * 0.32^0 * ( 1- 0.32 ) ^15
= 0.4477

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