Question

Assume the hold time of callers to a cable company is normally distributed with a mean of 4.7 minutes and a standard deviation of 0.9 minute. Determine the percent of callers who are on hold between 3.9 minutes and 6.2 minutes.

Answer 1

To find the percent of callers on hold between 3.9 and 6.2 minutes, calculate the Z-scores for each time and find the corresponding probabilities using the standard normal distribution. Subtract the lower probability from the higher to get the final percentage.

Step-by-step explanation:

To determine the percent of callers who are on hold between 3.9 minutes and 6.2 minutes for a normally distributed caller hold time with a mean of 4.7 minutes and a standard deviation of 0.9 minute, we need to calculate the Z-scores for each of the hold times and find the corresponding probability from the standard normal distribution table.

First, let's find the Z-score for 3.9 minutes:
Z = (X - μ) / σ
Where X is the hold time, μ (mu) is the mean, and σ (sigma) is the standard deviation.
Z for 3.9 minutes = (3.9 - 4.7) / 0.9 = -0.89

Now, calculate the Z-score for 6.2 minutes:
Z for 6.2 minutes = (6.2 - 4.7) / 0.9 = 1.67

Using the standard normal distribution table, find the probability for Z = -0.89 and Z = 1.67. The percent of callers between these Z-scores would be the probability for Z = 1.67 minus the probability for Z = -0.89.