Question

The number of reviews for each of the local restaurants is counted. The results are normally distributed with a mean of 25 and a standard deviation of 4.
What percentage of restaurants have between 20 and 25 reviews?

10.6%
50.0%
16.0%
39.4%

Answer 1

Using the normal distribution, it is found that 39.4% of restaurants have between 20 and 25 reviews.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by $\mu = 25, \sigma = 4$.

The proportion of restaurants that have between 20 and 25 reviews is the p-value of Z when X = 25 subtracted by the p-value of Z when X = 20, hence:

X = 25:

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

$Z = \frac{25 - 25}{4}$

Z = 0.

Z = 0 has a p-value of 0.5.

X = 20:

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

$Z = \frac{20 - 25}{4}$

Z = -1.25.

Z = -1.25 has a p-value of 0.106.

0.5 - 0.105 = 0.394 = 39.4%.

Hence, 39.4% of restaurants have between 20 and 25 reviews.

More can be learned about the normal distribution at brainly.com/question/24663213