A study of long distance phone calls made from the corporate offices of Pepsi Bo

Question

A study of long distance phone calls made from the corporate offices of Pepsi Bottling Group, Inc., in Somers, New York, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.2 minutes and the standard deviation was 0.50 minutes.

(a) What fraction of the calls last between 3.2 and 4 minutes?

(b) What fraction of the calls last more than 4 minutes?

(c) What fraction of the calls last between 4 and 4.5 minutes?

(d) What fraction of the calls last between 3 and 4.5 minutes?

Explanation / Answer

The general technique involves changing things into z-scores, and then looking up those numbers in a table. The way to do the z-scores:

P[a < X < b] = P[(a - mean)/st.dev. < z < (b - mean)/st.dev.] = P[z < (b - mean)/st.dev.] - P[z < (a - mean)/st.dev.]

For this problem, mean is 3.2 and st. dev. is 0.50.

a. P[3.2 < x < 4] = 0.4452.

b. P[4 < x] = 0.0548

c. P[4 < x < 4.5] = 0.0501.

d. P[3 < x < 4.5] = 0.6508.

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