Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 25, \sigma = 4[/tex].
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{25 - 25}{4}[/tex]
Z = 0.
Z = 0 has a p-value of 0.5.
X = 20:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20 - 25}{4}[/tex]
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 https://brainly.com/question/24663213
