91Ó°ÊÓ

IQ score,$x=125or150$

Mean,$\mathrm{μ}=110$

Standard deviation,$\mathrm{σ}=25$

The formula used:$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

Calculate the $z−$ score,

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}=\frac{125−110}{25}=0.60$

Or

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}=\frac{150−110}{25}=1.60$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with 1.6 and the column that starts with .00 of the standard normal probability table for P (z < 1.60)

Or

See the row that starts with 0.6 and the column that starts with .00 of the standard normalprobability table for P (z < 0.60)

$$\begin{gathered} P(125<x<150)=P(0.60<z<1.60)=P(z<1.60)−P(z<0.60) \\ =0.9452−0.7257 \\ =0.2195 \\ =21.95\% \end{gathered}$$

Therefore,

$21.95\%$ of the people aged $20$ to $34$ have IQs between $125$ and $150$

Find the z − score corresponding to the probability of 98% (or 0.98) in the normal probability table of the appendix.

Note that

The closest probability would be 0.9798 which lies in row 2.0 and in column .05 of the normal probability table.

Then

The corresponding z − score,

$z=2.05$

Calculate the z − score,

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}=\frac{x−110}{25}$

Equate the above two expressions:

$\frac{x−110}{25}=2.05$

Multiply both sides by 25:

$x−110=51.25$

Add 110 to both sides:

$x−110+110=51.25+110$

That becomes

$x=161.25$

Therefore,

An individual have to score at least 161.25 to become eligible for MENSA membership.