statistics

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I have the following homework question. I also included the work done. Please advise if I am on the right track Thank you!!

"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%

  • statistics -

    Close. I get Z = 1.47, P = .0708 = 7.08%

  • statistics -

    so only 7.08 % of employees would be entitled to the bonus?

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