91Ó°ÊÓ

The Wechsler IQ test has normally distributed scores with mean 100 and standard deviation 15.

The eligibility of Mensa Membership requires a score in top 2%.

Let X be the random variable for scores of IQ test.

Then,

$$\begin{gathered} X~N\left(\mathrm{μ},{\mathrm{σ}}^2\right) \\ ~N\left(100,{15}^2\right) \end{gathered}$$

a.

Let the minimum Wechster IQ score that satisfiy the Mensa requirement be x and the corresponding z-score be z such that,

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}...\left(1\right)$

Also,

$$\begin{gathered} P\left(X>x\right)=0.02 \\ P\left(Z>z\right)=0.02 \\ 1-P\left(Z<z\right)=0.02 \\ P\left(Z<z\right)=0.98...\left(2\right) \end{gathered}$$

Using Excel function =NORM.INV(0.98,0,1) the related z-score is computed as 2.0538.

Substitute the z-score in equation (1) to find the value of x.

$$\begin{gathered} 2.0538=\frac{x-100}{15} \\ x=2.0538×15+100 \\ =130.807 \\ =131 \end{gathered}$$

Thus, the minimum Wechsler IQ score which satisfies the experiment is 131.

b.

Since the population of IQ scores is normally distributed, the sample distribution of means would also be normal.

Let be the sample mean distribution of samples of size 4(n).

The mean and standard deviation of $\stackrel{¯}{X}$are;

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{15}{\sqrt{4}} \\ =7.5 \end{gathered}$$

Thus, $\stackrel{¯}{X}~N\left(100,{7.5}^2\right)$.

The Z-score associated to 131 is,

$$\begin{gathered} z=\frac{\stackrel{¯}{X}-{\mathrm{μ}}_{\stackrel{¯}{X}}}{{\mathrm{σ}}_{\stackrel{¯}{X}}} \\ =\frac{131-100}{7.5} \\ =4.1333 \end{gathered}$$

The probability that the mean score is at least 131 is the right tailed area to 131 on the sample distribution curve. Mathematically,

$$\begin{gathered} P\left(\stackrel{¯}{X}>131\right)=P\left(Z>4.1333\right) \\ =1-P\left(Z<4.1333\right) \end{gathered}$$

Using the Excel function =NORM.DIST(4.1333,0,1), the left tailed probability is obtained as 0.999982.

$$\begin{gathered} P\left(\stackrel{¯}{X}>131\right)=1-P\left(Z<4.1333\right) \\ =1-0.9999821 \\ =0.0000179 \end{gathered}$$

Thus, the probability that the mean score is at least 131 is 0.0000179.

c.

It is known that 4 subjects have mean score of 132.

It is stated that individual scores of four of them satisfy the eligibility for membership.

The statement is false, as it is possible that individual scores are below 131 for each individual even if the mean is 132.

For example,

$$\begin{gathered} {\stackrel{¯}{x}}_1=132 \\ \frac{{∑}_{i=1}^{4}x}{4}=132 \\ \underset{i=1}{\overset{4}{∑}}x=528 \end{gathered}$$

In this case, one of the combination of individual scores that satisfy the criteria is 100,100,158,170. Here, two individual do not qualify for the membership.