Question

scores on the wechsler adult intelligence scale (a standard iq test) for the 20 - 34 age - group are approximately normally distributed with $mu = 110$ and $sigma = 25$.
(a) what percent of people aged 20 to 34 have iqs between 125 and 150? (round to 2 decimal places.)
(b) mensa is an elite organization that admits as members people who score in the top 3% on iq tests. what score on the wechsler adult intelligence scale would an individual aged 20 to 34 have to earn to qualify for mensa membership? (round to 2 decimal places.)

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 110$ and $\sigma = 25$.
For $x = 125$, $z_1=\frac{125 - 110}{25}=\frac{15}{25}=0.6$.
For $x = 160$, $z_2=\frac{160 - 110}{25}=\frac{50}{25}=2$.
We use the standard normal distribution table (z - table) to find the probabilities associated with these z - scores.
The probability corresponding to $z = 0.6$ from the z - table is $P(Z<0.6)=0.7257$.
The probability corresponding to $z = 2$ from the z - table is $P(Z < 2)=0.9772$.
The probability of $Z$ being between $0.6$ and $2$ is $P(0.6

Step2: Find the z - score for the top 3%

If we want to find the z - score for the top 3% of the distribution, we first note that the area to the left of this z - score is $1 - 0.03=0.97$.
Looking up the value $0.97$ in the standard normal distribution table, the corresponding z - score is approximately $z = 1.88$.
We then use the z - score formula $z=\frac{x-\mu}{\sigma}$ to solve for $x$.
Rearranging the formula gives $x=\mu+z\sigma$.
Substituting $\mu = 110$, $z = 1.88$, and $\sigma = 25$, we get $x=110+1.88\times25=110 + 47=157$.

Answer:

(a) $25.15\%$
(b) $157.00$