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

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([tex]\mu = 110,\sigma^{2} = 15^{2}[/tex])

The z-score probability distribution is given by ;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean score = 110

[tex]\sigma[/tex] = 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 [tex]\geq[/tex] [tex]x[/tex] ) = 0.02 {where [tex]x[/tex] is minimum score required by person}

P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq \frac{x-110}{15}[/tex] ) = 0.02

P(Z [tex]\geq \frac{x-110}{15}[/tex] ) = 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; [tex]\frac{x-110}{15} = 2.0537[/tex]

[tex]x-110=2.0537 \times 15[/tex]

[tex]x[/tex] = 110 + 30.806 = 140.81

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