Quiz #7 / 80 points
DIRECTIONS:Show your work toearn partial credit.
1.The management of a large insurance company believes that workers are more productive if they are happy with their jobs.To keep track of their 815 workers’ satisfaction, the company regularly conducts surveys.According to a recent survey, the mean job satisfaction score for all workers at this company was 13.25 (on a scale of 1 to 20) and the standard deviation was 1.6.Assume that the job satisfaction scores of workers are normally distributed. (56 points)
a)What is the probability that a randomly selected worker will have a job satisfaction score between 14 and 18.5?Interpret the meaning of your answer. (16 points)
Ans.
p(14<x<18.5)
= p((14-13.25)/1.6 < z < (18.5-13.25)/1.6)
= p(0.46875 < z < 3.28125)
= 0.3191
This means, about 31.91% of workers have job satisfaction between 14 and 18.5
b)A worker with a score of 9.0 or less is considered very unhappy with his/her job. Approximately how many workers are very unhappy with their jobs?(15 points)
Ans.
P(x<9.0)
= p(x<(9-13.25)/1.6)
= p(z < -2.65625)
= 0.00395
So, number of workers unhappy = 815*0.00395 ≈ 3
c)A worker with a score in the top 7% of scores is considered to be very happy with his/her job.What is the minimum score needed to be considered very happy? Round your answer to the nearest tenth.Which percentile have you just found? (15 points)
Ans.
For top 7%, z >1.4758
(x-13.25)/1.6 = 1.4758
Or, x = 13.25+1.6*1.4758 = 15.6
This is 93rd percentile.
d)A worker at this company with a score in the bottom 2.5% of scores is viewed as needing intensive job counseling with a job satisfaction score of ______or less.Fill in the blank. Round to the nearest tenth. (10 points)
Ans.
For bottom 2.5%, z < -1.96
(x -13.25)/1.6 = -1.96
Or, x = 13.25-1.6*1.96 = 10.1
2.A television manufacturer is studying television remote control usage.One of the criteria they are measuring is the distance at which people attempt to activate the television set with the remote control.They have discovered that the activation distances are normally distributed with a mean distance of 6 feet and a standard deviation of 2.75 feet.If a remote’s maximum range is 10 feet, what percentage of the time will users attempts to operate the remote outside of its operating limit?(12 points)
Ans.
P(x>10)
= p(z > (10-6)/2.75)
= p( z > 1.4545)
= 0.0729
So, about 7.29% times users would attempt to operate outside operating limit.
3. Scores for a civil service exam are normally distributed, with a standard deviation of 6.9.To be eligible for civil service employment, you must score in the top 5%.If the lowest score you can earn and still be eligible for employment is an 84, what is the mean score for the exam? Round your answer to the nearest tenth.(12 points)
Ans.
For top 5%, z > 1.6449
(84-µ)/6.9 = 1.6449
Or, µ = 84 - 1.6449*6.9 = 72.7