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 9.7 minutes and a standard deviation of 2.4 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.

(a) less than 10 minutes
(b) longer than 5 minutes
(c) between 8 and 15 minutes

Answer 1

Answer:

a) 89.44% probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be less than 10 minutes.

b) 97.50% probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be longer than 5 minutes.

c) 74.75% probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be between 8 and 15 minutes.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.7, \sigma = 2.4$

(a) less than 10 minutes

This is the pvalue of Z when X = 10. So

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

$Z = \frac{10 - 9.7}{2.4}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

89.44% probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be less than 10 minutes.

(b) longer than 5 minutes

This is 1 subtracted by the pvalue of Z when X = 5.

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

$Z = \frac{5 - 9.7}{2.4}$

$Z = -1.96$

$Z = -1.96$ has a pvalue of 0.0250

1 - 0.0250 = 0.9750

97.50% probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be longer than 5 minutes.

(c) between 8 and 15 minutes

This is the pvalue of Z when X = 15 subtracted by the pvalue of Z when X = 8.

X = 15

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

$Z = \frac{15 - 9.7}{2.4}$

$Z = 2.21$

$Z = 2.21$ has a pvalue of 0.9864

X = 8

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

$Z = \frac{8 - 9.7}{2.4}$

$Z = -0.71$

$Z = -0.71$ has a pvalue of 0.2389

0.9864 - 0.2389 = 0.7475

74.75% probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be between 8 and 15 minutes.