## Question 815:

1

For this question we use the normal distribution and formula for a z-score = (x-mean)/sd

1. For part a we want the area above the point in the normal curve using the z-score formula = (1500-1280)/420 = .52381 and the probability associated with this z-score from a normal table or excel formula =NORMSDIST() = is .6998, so 1-.6998 = .3002055 or about 30% or 1/3 of employees cost more than \$1500
2. For between area questions we calculate z-scores for both data points and subtract the smaller from the larger. The smaller was .6998 and the larger is the z-score 1.7143 = .95676 - .6998 = .2569 or about ¼ of the employees cost between \$1500 and \$2k per year.
3. Having no expense is an x of 0 in the formula = (0-1280)/420 = -3.047, so technically the probability for exactly 0 is 0 (from the law of the continuous probability distribution), but if we say the probability from 0 to –infinity, it's .0012, the probability for -3.047 so less than 1/100 would have \$0 or less.
4. The highest 10% or 90th percentile and above requires 1st finding the z-score for .90 = 1.28.
1. so setting the equation (x-1280)/420 = 1.28 and solve for x
2. x-1280 = 538.2517
3. x = 1818.252
4. So \$1818.252 and above were the most expensive 10% of employees.