Question

question 7 of 10 > 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 $sigma = 25$. (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 2% 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?

Explanation:

Step1: Standardize the values

We use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 110$ and $\sigma=25$.
For $x = 125$, $z_1=\frac{125 - 110}{25}=\frac{15}{25}=0.6$.
For $x = 150$, $z_2=\frac{150 - 110}{25}=\frac{40}{25}=1.6$.

Step2: Find the probabilities

We use the standard normal distribution table (z - table).
$P(0.6From the z - table, $P(Z < 1.6)=0.9452$ and $P(Z < 0.6)=0.7257$.
So $P(0.6

Step3: Convert to percentage

The percentage is $0.2195\times100 = 21.95\%$.

Step4: For part (b)

We want to find the z - score corresponding to the top 2% or the 98th percentile.
Looking up in the z - table, the z - score corresponding to a cumulative probability of 0.98 is approximately $z = 2.05$.

Step5: Solve for x

Using the z - score formula $z=\frac{x-\mu}{\sigma}$ and re - arranging for $x$, we get $x=\mu+z\sigma$.
Substituting $\mu = 110$, $z = 2.05$ and $\sigma = 25$, we have $x=110+2.05\times25=110 + 51.25=161.25$.

Answer:

(a) 21.95%
(b) 161.25