IQ test scores Scores on the Wechsler Adult Intelligence Scale (a standard IQ test) for the $20$to $34$age group are approximately Normally distributed with $\mu =110$and $\sigma =25$. For each part, follow the fourstep process.

(a) At what percentile is an IQ score of $150$?

(b) What percent of people aged $20$to $34$have IQs between $125$and $150$?

(c) MENSA is an elite organization that admits as members people who score in the top $2\%$on IQ tests. What score on the Wechsler Adult Intelligence Scale would an individual have to earn to qualify for MENSA membership?

Mean$\mu =110$

Standard deviation$\sigma =25$

Students' IQ scores are variable $x$. It has a normal distribution with $\mu =110and\sigma =25$.

A low IQ score is one with less than$150$points.

In the graph below, you can see how many people have an IQ score below $150:$

Following are the corresponding $z$values calculated for$x=150$

$$\begin{gathered} z=\frac{150-110}{25} \\ =1.6 \end{gathered}$$

With a standard normal table, we can see that the proportion of observations with less than$1.6$is$0.9452$or approximately $94.5\%$

Hence, an IQ score of 150 is approximately at the${95}^{th}$percentile.

Let x be the IQ scores of students. The variable $x$has a Normal distribution with $\mu =110and\sigma =25$. We want the proportion of IQ scores that are between$125and150$.

As shown in the graph below, a majority of IQ scores fall between $125$and $150$

From (a) part $x=150andz=1.6$

We have$x=125$and the z value can be find below:

$$\begin{gathered} z=\frac{125-110}{25} \\ =0.6 \end{gathered}$$

A proportion between $0.6$and $1.6$ is equal to a proportion between $1125$and $150.$

The area between $0.6and1.6is0.9452-0.7257=0.2195$or$22\%$by using standard normal table,

So, approximately$22\%$of people aged$22-34$have IQ scores between $12and150$

Let x be the IQ scores of students. The variable x has a Normal distribution with $\mu =110and\sigma =25$we have to find out the score such that$98\%$of people aged$20-34$have an IQ score that is smaller

The graph below shows the${98}^{th}$percentile of the IQ score:

Thelocalid="1649931146189" ${98}^{th}$percentile for the standard Normal distribution is approximate to localid="1649931131184" $2.05$by using the standard normal table,

So, the IQ score oflocalid="1649931138907" ${98}^{th}$percentile can be found by:

localid="1649997385362" $$\begin{gathered} 2.05=\frac{x-110}{25} \\ x=161.25 \end{gathered}$$

MENSA membership is only open to people with scores of localid="1649931178016" $162$or higher.