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

Answer:

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

Step-by-step 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 = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ $\geq \frac{x-110}{15}$ ) = 0.02

             P(Z $\geq \frac{x-110}{15}$ ) = 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;         $\frac{x-110}{15} = 2.0537$

                          $x-110=2.0537 \times 15$  

                              $x$ = 110 + 30.806 = 140.81

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