Question

In order to qualify for letter sorting l, applicants are given a speed-reading test. The scores are normally distributed, with a mean of 80 and a standard deviation of 8. If only the top 15% of the applicants are selected, find the score needed to qualify

Answer 1

Using the normal distribution, it is found that the score needed to qualify is of 88.32.

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, respectively, by $\mu = 80$ and $\sigma = 8$.

The top 15% of the applicants are selected, hence the score needed to qualify is the 85th percentile, which is X when Z = 1.04, hence:

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

$1.04 = \frac{X - 80}{8}$

X - 80 = 1.04(8)

X = 88.32.

The score needed to qualify is of 88.32.

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