Question

A particular type of 4th grade Achievement Test provides overall scores that are normally distributed with a mean of 50 and a standard deviation of 10. One state wants to allow all students with scores in the top 3% into a special advanced program.
a. What will be the minimum score required to be admitted into this program?
b. One state wants to allow all students with scores in the top 38 into a special advanced program. What will be the minimum score required to be admitted into this program?

Answer 1

Answer:

a) The minimum score required to be admitted into this program will be 68.8.

b) The minimum score required to be admitted into this program will be 53.05.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 10$

a. What will be the minimum score required to be admitted into this program?

Top 3%, so the minimum score is X when Z has a pvalue of 1-0.03 = 0.97. So X when Z = 1.88.

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

$1.88 = \frac{X - 50}{10}$

$X - 50 = 10*1.88$

$X = 68.8$

The minimum score required to be admitted into this program will be 68.8.

b. One state wants to allow all students with scores in the top 38 into a special advanced program. What will be the minimum score required to be admitted into this program?

Top 38%, so the minimum score is X when Z has a pvalue of 1-0.32 = 0.68. So X when Z = 0.305.

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

$0.305 = \frac{X - 50}{10}$

$X - 50 = 10*0.305$

$X = 53.05$

The minimum score required to be admitted into this program will be 53.05.