"A recent study determined that the distance employees at a pharmaceutical company in New Jersey commute to work each way is normally distributed with a mean equal to 13.4 miles and a standard deviation of 3.6 miles. Management has decided to give a year-end bonus to employees who commute large distances and have decided to give the bonus to any employee who commutes over 18.7 miles. Based on this information, approximately 13.3 percent of the employees will get the year-end bonus.

Consider this situation and respond as to whether you agree with Management's conclusion that 13.3 percent of the employees will get the year-end bonus. Provide your rationale if you disagree with Management and provide the correct percent."

WORK DONE :

z(18.7) = (18.7-13.4)/3.6 = 1.5143

P(x > 18.7) = P(z > 1.5143) = 0.0650 or 6.5%

I believe that only 6.5% will get the bonus and not 13.3. Is this correct?

Based on the calculations you have done, it seems that you believe only 6.5% of the employees will get the year-end bonus, which differs from Management's conclusion of 13.3%.

Your calculation to find the z-score for a commute distance of 18.7 miles appears to be correct:

z(18.7) = (18.7 - 13.4) / 3.6 = 1.5143

Next, you correctly found the probability that a randomly selected employee commutes more than 18.7 miles using the standardized normal distribution (z-distribution) table:

P(x > 18.7) = P(z > 1.5143) = 0.0650 or 6.5%

Therefore, based on your calculations, it seems that approximately 6.5% of the employees at the pharmaceutical company will commute over 18.7 miles and receive the year-end bonus.

If you disagree with Management's conclusion of 13.3%, it's important to consider that your calculation assumes a normal distribution for the commute distances, which is based on the given information. However, it's possible that the actual distribution of commute distances is not perfectly normally distributed. It would be useful to conduct further analysis or gather more data to determine if the assumption of a normal distribution is appropriate in this scenario.