Question

A person must score in the upper 2% of the population on an admissions test to qualify for membership in society catering to highly intelligent individuals. If test scores are normally distributed with a mean of 110 and a standard deviation of 15, what is the minimum score a person must have to qualify for the society

Answer 1

The minimum score a person must have to qualify for the society is 140.81.

Explanation:

We are given that a person must score in the upper 2% of the population on an admissions test to qualify for membership in society catering to highly intelligent individuals.

Also, test scores are normally distributed with a mean of 110 and a standard deviation of 15.

Let X = test scores

SO, X ~ N(
`\mu = 110,\sigma^(2) = 15^(2)`)

The z-score probability distribution is given by ;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = mean score = 110

`\sigma` = standard deviation = 15

Now, the minimum score a person must have to qualify for the society so that his score is in the top 2% is given by ;

P(X
`\geq`
`x` ) = 0.02 {where
`x` is minimum score required by person}

P(
`(X-\mu)/(\sigma)`
`\geq (x-110)/(15)` ) = 0.02

P(Z
`\geq (x-110)/(15)` ) = 0.02

Now, in z table we will find out that critical value of X for which the area is in top 2%, which comes out to be 2.0537

This means;
`(x-110)/(15) = 2.0537`

`x-110=2.0537 * 15`

`x` = 110 + 30.806 = 140.81

Therefore, the minimum score a person must have to qualify for the society is 140.81.