Question

A small engine shop receives an average of repair calls per​ hour, with a standard deviation of . What is the mean and standard deviation of the number of calls it receives for ​-hour ​day? What, if​ anything, did you​ assume?

Answer 1

Answer:

Assuming normal distribution, the mean number of calls for a n-hour day is of $m = n\mu$, in which $\mu$ is the mean number of calls per hour, and the standard deviation is $s = \sqrt{n}\sigma$, in which $\sigma$ is the standard deviation of the number of calls per hour.

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.

N-instances of a normal variable:

For n-instances of normal variable, the mean of the distribution is: $m = n\mu$, and the standard deviation is $s = \sqrt{n}\sigma$

What is the mean and standard deviation of the number of calls it receives for ​n-hour ​day?

Assuming normal distribution, the mean number of calls for a n-hour day is of $m = n\mu$, in which $\mu$ is the mean number of calls per hour, and the standard deviation is $s = \sqrt{n}\sigma$, in which $\sigma$ is the standard deviation of the number of calls per hour.