Question

Suppose the daily customer volume at a call center has a normal distribution with mean 5,500 and standard deviation 1,000. What is the probability that the call center will get between 4,800 and 5,000 calls in a day

Answer 1

Answer:

0.0665 = 6.65% probability that the call center will get between 4,800 and 5,000 calls in a day.

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.

Mean 5,500 and standard deviation 1,000.

This means that $\mu = 5500, \sigma = 1000$

What is the probability that the call center will get between 4,800 and 5,000 calls in a day?

This is the p-value of Z when X = 5000 subtracted by the p-value of Z when X = 4800. So

X = 5000

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

$Z = \frac{5000 - 5500}{1000}$

$Z = -0.5$

$Z = -0.5$ has a p-value of 0.3085.

X = 4800

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

$Z = \frac{4800 - 5500}{1000}$

$Z = -0.7$

$Z = -0.7$ has a p-value of 0.2420.

0.3085 - 0.2420 = 0.0665

0.0665 = 6.65% probability that the call center will get between 4,800 and 5,000 calls in a day.