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4. (40 Pts) The Vacation Times website rates recreational vehicle campgrounds
using integers from 0 to 15. Last year they rated over 1,000 campsites. The
ratings were normally distributed with mean 7.6 and standard deviation 1.7.
a. How high would a campsite's rating have to be for it to be considered in
the top 10% of rated campsites? Round to the nearest hundredth.
b. Find the z-score for a rating of 5. Round to the nearest hundredth.

Respuesta :

Using the normal distribution, it is found that:

a) The rating would have to be of at least 9.78.

b) The z-score for a rating of 5 is of -1.53.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given by:

[tex]\mu = 7.6, \sigma = 1.7[/tex]

For item a, we have to find the 90th percentile, which is X when Z = 1.28, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

1.28 = (X - 7.6)/1.7

X - 7.6 = 1.28 x 1.7

X = 9.78.

The rating would have to be of at least 9.78.

For item b, we have to find Z when X = 5, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (5 - 7.6)/1.7

Z = -1.53.


The z-score for a rating of 5 is of -1.53.

More can be learned about the normal distribution at https://brainly.com/question/24808124

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