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8 November, 02:20

Assume the hold time of callers to a cable company is normally distributed with a mean of 3.53.5 minutes and a standard deviation of 0.20.2 minute. Determine the percent of callers who are on hold between 3.43.4 minutesminutes and 3.83.8 minutes.

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  1. 8 November, 05:46
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    We calculate z-scores associated with x = 3.4 and x = 3.8.

    For x = 3.4, z = (3.4 - 3.5) / (0.2) = - 0.5

    For x = 3.8, z = (3.8 - 3.5) / (0.2) = 1.5

    Then we calculate the probability that - 0.5 < z < 1.5 from z-tables.

    P (-0.5 < z < 1.5) = 0.6247 = 62.47% of callers on hold between 3.4-3.8 minutes.
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