Opening up the monthly cell phone bill can be alarming due to the uncertainty of the bill. suppose the amount of minutes used per month is distributed normally with a mean of 700 minutes and a standard deviation of 120 minutes. what is the probability that more than 940 minutes were used?

Accepted Solution

Given the mean of 700 min. and the std. dev. of 120 min., we need to find the z-score for 940 minutes and then the area under the std. normal curve to the right of that z-score.
                                                        940 - 700
The applicable z score here is z = ----------------- = 2.  What is the area under
the curve and to the right of 2 std. deviations from the mean?

You could find the answer from a table of z-scores, or you could take advantage of the empirical rule.  The empirical rule states that 95% of the normally distributed data set lies within 2 std. dev. of the mean.

The area to the right of z=2 is 0.5 (half the area under the std. normal curve) less 0.95/2, that is, less 0.475.

Subtracting 0.475 from 0.500 yields 0.025 (answer).  2.5% of the time, the monthly phone bill will exceed 940, or 2 std. dev. above the mean.