The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 65 and a standard deviation of 8. Using the Empirical Rule rule, what is the approximate percentage of lightbulb replacement requests numbering between 41 and 65? Do not enter the percent symbol.

We want to approximate the following probability with a %:

And we can find the number of deviations from the mean for the limits with this formula called the z score:

And using the empirical rule we know that within 3 deviation from the mean we have 0.997 of the area so then for the middle of the area we need to have 0.4985

So we can approximate the probability required as:

And the percent would be 49.85

Step-by-step explanation:

Previous concepts

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the number of daily requests to replace lightbulbs.

From the problem we have the mean and the standard deviation for the random variable X.

So we can assume

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

•The probability of obtain values within one deviation from the mean is 0.68

•The probability of obtain values within two deviation's from the mean is 0.95

•The probability of obtain values within three deviation's from the mean is 0.997

Solution to the problem

We want to approximate the following probability with a %:

And we can find the number of deviations from the mean for the limits with this formula called the z score:

And using the empirical rule we know that within 3 deviation from the mean we have 0.997 of the area so then for the middle of the area we need to have 0.4985

So we can approximate the probability required as:

And the percent would be 49.85