Final answer:
Approximately 51.22% of callers are on hold between 2.3 and 3.8 minutes, calculated using z-scores and the standard normal distribution for a hold time with a mean of 3.6 minutes and a standard deviation of 0.9 minutes.
Step-by-step explanation:
To determine the percent of callers who are on hold between 2-3 and 3.8 minutes when the hold times are normally distributed with a mean of 3.6 minutes and a standard deviation of 0.9 minutes, we can use the standard normal distribution and z-scores.
First, we calculate the z-scores for 2-3 minutes and 3.8 minutes using the formula:
Z = (X - mean) / standard deviation
For 2.3 minutes:
Z = (2.3 - 3.6) / 0.9 = -1.44
For 3.8 minutes:
Z = (3.8 - 3.6) / 0.9 = 0.22
Next, we find the probability corresponding to these z-scores using the standard normal distribution table or a calculator. For Z = -1.44, the probability is approximately 0.0749, and for Z = 0.22, it is approximately 0.5871. To find the percentage of callers between these two times, take the difference of these probabilities:
Percentage = (0.5871 - 0.0749) * 100%
Percentage = 51.22%
Therefore, approximately 51.22% of callers are expected to be on hold for between 2.3 and 3.8 minutes.