Question

Employee Satisfaction Rates: A recent poll suggests that 48% of Americans are satisfied with their job. You have a company with 226 employees and a poll suggests that 89 of them are satisfied (about 39%). (a) Assume the 48% general satisfaction rate is accurate. In all random samples of 226 Americans, what is the mean and standard deviation for the number of people satisfied with their job? Round both answers to 1 decimal place. \muμ = \sigmaσ = (b) Convert the 89 out of 226 satisfied employees at your company to a z-score. Round your answer to 2 decimal places. z = ? (c) Using the normal approximation to the binomial distribution, what is the probability of getting 89 or fewer satisfied employees in a randomly selected group of 226? Round your answer to 4 decimal places. P(x ≤ 89) = (d) What conclusions may you be able to make about job satisfaction at your company?

Answer #1

a)

here mean of distribution=μ=np= | 108.5 |

and standard deviation σ=sqrt(np(1-p))= | 7.5 |

b)

z score =(X-mean)/stadnard deviation =(89-108.5)/7.5 = -2.60 (please try -2.59 if this comes wrong)

c) probability of getting 89 or fewer satisfied employees in a randomly selected group of 226

=P(Z<-2.60) =0.0047 (please try 0.0048 if this comes wrong)

d)

since probability of getting such or more extreme result is highly unusual, we doubt that % who are satisfied is lower than 48%.

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