At a large department store, the average number of years of service was 13.5 with a standard deviation of 6.2?

At a large department store, the average number of years of service was 13.5 with a standard deviation of 6.2 years. If an employee is picked at random, what is the probability that the employee has worked for the company for over 14 years?











a. 0.0318





b. 0.0852





c. 0.1875





d. 0.2862





e. 0.4679





f. 0.6566





g. none of these

At a large department store, the average number of years of service was 13.5 with a standard deviation of 6.2?
To be employed for more than 14 years means the employee is employed for 0.5 years greater than the average. 0.5 represents 0.0806 standard deviations from the mean. Looking up the z score you get .0321 (interpolating). This means that there are .5321 below 14 years or .4679 at 14 years or more. So the answer is e.
Reply:z = (X - μ) / σ





z = (14 - 13.5) / 6.2 ≈ 0.0806


P(z %26gt; 0.0806) ≈ 0.4679





The answer is e.