answer to quest

Question

Suppose the mean income of firms in the industry for a year is 65

million dollars with a standard deviation of 1

million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 66

million dollars?

Answer 1

To answer this question, you need to know how many standard deviations $66M is away from the mean of $65M. Find the difference and divide by the standard deviation:

(66M - 65M)/1M = 1.00

Since you are given that the incomes are normally distributed, this figure, 1.00 is a z-score. (By convention we refer to quantiles of the standard normal distribution as z-scores.) You can find the probability by looking that z-score up in a table of the standard normal distribution, or by using the NORM.DIST function in Excel. You want the probability in the left tail because the question asks for the probability that income is less than $66M. Most tables display the left tail probability.