Question

ACT scores are normally distributed and from 2015 to 2017, the mean was 20.9 with a standard deviation of 5.6. A perfect score is 36. What percentage of the testers scored a perfect score?

Question 4 options:

0.46%


0.35%


99.65%


99.64%

Answer 1

Using the normal distribution, it is found that 0.35% of the testers scored a perfect score.

Normal Probability Distribution

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

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

• 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.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 20.9, \sigma = 5.6$

The proportion of scores above 36(perfect scores) is given by one subtracted by the p-value of Z when X = 36, hence:

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

$Z = \frac{36 - 20.9}{5.6}$

Z = 2.7

Z = 2.7 has a p-value of 0.9965.

1 - 0.9965 = 0.0035 = 0.35.%.

Hence, 0.35% of the testers scored a perfect score.

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

#SPJ1