Question

PREVIOUS ANSWERS MY NOTES Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minutes. An operator in the call center is required to answer 75 calls each day. Assume the call times are independent. What is the expected total amount of time in minutes the operator will spend on the calls each day

Answer 1

Answer:

The expected total amount of time in minutes the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

n-values of a normal variable:

For the sum of a sample of n values, the mean is of $M = n\mu$ and the standard deviation is of $s = \sigma\sqrt{n}$

Average 2.8 minutes

This means that $\mu = 2.8$

75 calls each day.

This means that $n = 75$

What is the expected total amount of time in minutes the operator will spend on the calls each day?

$M = n\mu = 75(2.8) = 210$

The expected total amount of time in minutes the operator will spend on the calls each day is of 210 minutes.