Chapter 2.2, Problem 67E

(a)

To determine

Percentage of the people aged 20 to 34 having IQs between 125 and 150.

Answer to Problem 67E

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

Explanation of Solution

Given information:

IQ score, $x=125\text{or}150$

Mean, $\mu =110$

Standard deviation, $\sigma =25$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{125-110}{25}=0.60$

Or

  $z=\frac{x-\mu }{\sigma }=\frac{150-110}{25}=1.60$

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 $P\left(z<1.60\right)$ .

Or

See the row that starts with 0.6 and the column that starts with .00 of the standard normal probability table for $P\left(z<0.60\right)$ .

  $\begin{array}{l}P\left(125<x<150\right)=P\left(0.60<z<1.60\right)\\ =P\left(z<1.60\right)-P\left(z<0.60\right)\\ =0.9452-0.7257\\ =0.2195\\ =21.95\%\end{array}$

Therefore,

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

(b)

To determine

Score to be earned by an individual aged 20 to 34 on the Wechsler Adult intelligence scale to qualify for MENSA membership.

Answer to Problem 67E

An individual requires score of at least 161.25 to qualify for MENSA membership.

Explanation of Solution

Given information:

Mean, $\mu =110$

Standard deviation, $\sigma =25$

2% (or 0.02) is greater than z .

98% (or 0.98) is smaller than z .

Calculations:

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

Note that

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

Then

The corresponding z − score,

  $z=2.05$

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{x-110}{25}$

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.