Question

The scores on an national exam for 11th grade is are normally distributed with and .

Which of the following statements most precisely describes the probability that a randomly chosen 11th grader scored 90 or more on this test?

Answer 1

Using the normal distribution, it is found that there is a 0.0174 = 1.74% probability that a randomly chosen 11th grader scored 90 or more on this test.

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.

Researching the problem on the internet, it is found that the mean and the standard deviation are given by, respectively, $\mu = 71, \sigma = 9$.

The probability that a randomly chosen 11th grader scored 90 or more on this test is one subtracted by the p-value of Z when X = 90, hence:

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

$Z = \frac{90 - 71}{9}$

Z = 2.11

Z = 2.11 has a p-value of 0.9826.

1 - 0.9826 = 0.0174.

0.0174 = 1.74% probability that a randomly chosen 11th grader scored 90 or more on this test.

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