Question

At a costumer service call center for a large company, the number of calls received per hour is normally distributed with a mean of 150 calls and a standard deviation of 5 calls. What is the probability that during a given hour of the day there will be between 146 calls and 163 calls, to the nearest thousandth

Answer 1

We have been given that at a customer service call center for a large company, the number of calls received per hour is normally distributed with a mean of 150 calls and a standard deviation of 5 calls. We are asked to find the probability that during a given hour of the day there will be between 146 calls and 163 calls.

First of all, we will use z-score formula to find z-score corresponding to 146 and 163.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{146-150}{5}$

$z=\frac{-4}{5}$

$z=-0.8$

$z=\frac{163-150}{5}$

$z=\frac{13}{5}$

$z=2.6$

Our next step is to find percentage of data scores falls between both z-scores.

$P(-0.8<z<2.6)=P(z<2.6)-P(z<-0.8)$

Using normal distribution table, we will get:

$P(-0.8<z<2.6)=0.99534-0.21186$

$P(-0.8<z<2.6)=0.78348$

Let us convert our answer into percentage.

$0.78348\times 100\%=78.348\%$

Upon rounding our answer to nearest thousandths, we will get:

$78.348\%\approx 78.35\%$

Therefore, the probability that during a given hour of the day there will be between 146 calls and 163 calls is approximately $78.35\%$.

Answer 2

Answer:

0.624

Step-by-step explanation:

DeltaMath