Question

use the normal model n(100,16) describing iq scores to answer the following. a) what percent of peoples iqs are expected to be over 85? b) what percent of peoples iqs are expected to be under 90? c) what percent of peoples iqs are expected to be between 120 and 136? a) approximately % of peoples iqs are expected to be above 85. (round to one decimal place as needed.)

Explanation:

Step1: Calculate the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 100$ is the mean and $\sigma=\sqrt{16}=4$ is the standard deviation.

Step2: Calculate z - score for part a

For $x = 85$, $z=\frac{85 - 100}{4}=\frac{- 15}{4}=-3.75$.
We want $P(X>85)$, which is $1 - P(X\leq85)$. Looking up $z=-3.75$ in the standard - normal table, $P(Z\leq - 3.75)\approx0.0001$. So $P(X>85)=1 - 0.0001 = 0.9999\approx99.99\%\approx100.0\%$.

Step3: Calculate z - score for part b

For $x = 90$, $z=\frac{90 - 100}{4}=\frac{-10}{4}=-2.5$.
Looking up $z=-2.5$ in the standard - normal table, $P(Z\leq - 2.5)=0.0062$. So approximately $0.6\%$ of people's IQs are expected to be under 90.

Step4: Calculate z - scores for part c

For $x = 120$, $z_1=\frac{120 - 100}{4}=\frac{20}{4}=5$.
For $x = 136$, $z_2=\frac{136 - 100}{4}=\frac{36}{4}=9$.
Since the standard - normal table usually only goes up to $z = 3.49$ and $P(Z\leq5)\approx1$ and $P(Z\leq9)\approx1$, $P(120

Answer:

a) 100.0%
b) 0.6%
c) 0%