A major department store has determined that its customers c

A major department store has determined that its customers charge an average of $500 per month, with a standard deviation of $80. Assume the amounts of charges are normally distributed.

a. What percentage of customers charges more than $380 per month?

b. What percentage of customers charges less than $340 per month?

c. What percentage of customers charges between $644 and $700 per month?

Solution

You will use the z-score formula and table to solve these.

The formula is (X-mean)/SD

a. (380-500)/80=-1.5. Looking at a z-table (which should be found in the back of your statistics textbook or else you can google it), that converts to .0668. That means that 6.68% of the customers charge LESS than 380 a month. Subract that from 1 and you get .9332, or 93.32% of customers.

b. (340-500)/80=-2. A z-table gives you .0228, and since we want the percent that is less, it is 2.28%. (A traditional z-table score gives you the percentage of people who are lower then that standard deviation).

c.If you have two numbers, you have to do two formulas and subtract the difference.

(644-500)/80=1.8 and (700-500)/80=2.5. .9641 and .9938. The difference between the two of those is .0297, therefore 2.97% of people spent between 644 and 700 dollars.