Employee Satisfaction Rates: A recent poll suggests that 48% of Americans are sa
ID: 3230712 • Letter: E
Question
Employee Satisfaction Rates: A recent poll suggests that 48% of Americans are satisfied with their job. You have a company with 214 employees and a poll suggests that 85 of them are satisfied (about 40%).
(a) Assume the 48% general satisfaction rate is accurate. In all random samples of 214 Americans, what is the mean and standard deviation for the number of people satisfied with their job? Round both answers to 1 decimal place.
(b) Convert the 85 out of 214 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 85 or fewer satisfied employees in a randomly selected group of 214? Round your answer to 4 decimal places.
P(x 85) =
(d) Which statement best describes the situation at your company?
There is no chance that the low satisfaction rate is due to random variation.
Since the probability of getting a sample of 214 people with fewer satisfied employees than you have is so small, this suggests that something is unusual with your company and/or your employees.
Having only 85 out of 214 employees being satisfied is not particularly unusual.
= =Explanation / Answer
a) mean =np=214*0.48=102.72
std deviation =(np(1-p))1/2 =7.3085
b)Z=(X-mean)/std deviation =(85.5-102.72)/7.3085)=-2.356 (by applying continutiy correction
c) P(X<=85)=P(Z<-2.356)=0.0092
Since the probability of getting a sample of 214 people with fewer satisfied employees than you have is so small, this suggests that something is unusual with your company and/or your employees.
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