Question

The Customer Service Center in a large New York department store has determined that the amount of time spent with a customer about a complaint is normally distributed, with a mean of 8.9 minutes and a standard deviation of 2.8 minutes. What is the probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be as follows. (Round your answers to four decimal places.)
(a) less than 10 minutes


(b) longer than 5 minutes


(c) between 8 and 15 minutes

Answer 1

Using the normal distribution, the probabilities are given as follows:

a) 0.6517 = 65.17%

b) 0.0823 = 8.23%.

c) 0.6109 = 61.09%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 8.9, \sigma = 2.8$

Item a:

The probability is the p-value of Z when X = 10, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 8.9}{2.8}$

Z = 0.39

Z = 0.39 has a p-value of 0.6517.

0.6517 = 65.17% probability that the time is less than 10 minutes.

Item b:

The probability is the one subtracted by the p-value of Z when X = 5, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 8.9}{2.8}$

Z = -1.39

Z = -1.39 has a p-value of 0.0823.

0.0823 = 8.23% probability that the time is more than 5 minutes.

Item c:

The probability is the p-value of Z when X = 15 subtracted by the p-value of Z when X = 8, hence:

X = 15:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{15 - 8.9}{2.8}$

Z = 2.18

Z = 2.18 has a p-value of 0.9854.

X = 8:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8 - 8.9}{2.8}$

Z = -0.32

Z = -0.32 has a p-value of 0.3745.

0.9854 - 0.3745 = 0.6109 = 61.09%.

More can be learned about the normal distribution at brainly.com/question/4079902

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