Question

at a customer service call center for a large company, the number of calls received per hour is normally distributed with a mean of 190 calls and a standard deviation of 10 calls. what is the probability that during a given hour of the day there will be more than 192 calls, to the nearest thousandth?
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Explanation:

Step1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value from the dataset, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given that $\mu = 190$, $\sigma=10$, and $x = 192$.
Substitute these values into the formula: $z=\frac{192 - 190}{10}=\frac{2}{10} = 0.2$

Step2: Find the probability $P(Z>0.2)$

We know that the total area under the standard normal curve is 1. And $P(Z>z)=1 - P(Z\leq z)$.
We can look up the value of $P(Z\leq0.2)$ in the standard normal distribution table (or use a calculator with a normal - distribution function). The value of $P(Z\leq0.2)\approx0.5793$ (from standard normal tables or calculator).
Then $P(Z > 0.2)=1 - 0.5793=0.4207$

Answer:

0.421