Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]M = n\mu[/tex] and the standard deviation is of [tex]s = \sigma\sqrt{n}[/tex]
Average 2.8 minutes
This means that [tex]\mu = 2.8[/tex]
75 calls each day.
This means that [tex]n = 75[/tex]
What is the expected total amount of time in minutes the operator will spend on the calls each day?
[tex]M = n\mu = 75(2.8) = 210[/tex]
The expected total amount of time in minutes the operator will spend on the calls each day is of 210 minutes.
