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

The expected total amount of time 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 and standard deviation , the z-score of a measure X is given by:

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 normal variable:

Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is and the standard deviation is

Calls to a customer service center last on average 2.8 minutes.

This means that

75 calls each day.

This means that

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

This is M, so:

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

step-by-step explanation:

(s)he  is right it would be 5.5% because it is smaller than 15.5% or anything smaller than 15.5% in general