223k views
2 votes
assume the hold time of callers to a cable company is normally distributed with a mean of 3.6 and a standard a deviation of .9 minute. determine the percent of callers who are on hold between 2-3 and 3.8 minutes

User Metalgear
by
8.4k points

1 Answer

1 vote

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.

User Troy Howard
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.