I need a bit more help
2 answers:
You might be interested in
Answer:
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.