Question

On a standardized test, Jose scored an 87 and Mia scored a 74. The distribution of test scores for the population is normal with a population mean of 76.28 and a population standard deviation of 7.02.

What percentage of people received a higher score than Jose?

Answer 1

Using the normal distribution, it is found that 6.3% of people received a higher score than Jose.

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 by, respectively: $\mu = 76.28, \sigma = 7.02$.

The proportion of people who received a higher score than Jose is one subtracted by the p-value of Z when X = 87, hence:

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

$Z = \frac{87 - 76.28}{7.02}$

$Z = 1.53$

$Z = 1.53$ has a p-value of 0.937.

1 - 0.937 = 0.063.

6.3% of people received a higher score than Jose.

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