Question

question 4
3 pts
iq test scores are standardized to produce a normal distribution with a mean of μ = 100 and a standard deviation of σ = 15. what is the proportion of the population in each of the following iq categories:
genius or near genius: iq over 140
z = select : p = select
very superior intelligence: iq over 120
z = select : p = select

Explanation:

Step1: Calculate z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation.

Step2: Calculate z - score for $x = 140$

For $x = 140$, $\mu=100$, $\sigma = 15$, we have $z=\frac{140 - 100}{15}=\frac{40}{15}\approx2.67$.

Step3: Find proportion for $z = 2.67$

We want $P(X>140)$, which is equivalent to $P(Z > 2.67)$. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. Looking up $P(Z\leq2.67)$ in the standard - normal table, we find $P(Z\leq2.67)\approx0.9962$. So $P(Z > 2.67)=1 - 0.9962 = 0.0038$.

Step4: Calculate z - score for $x = 120$

For $x = 120$, $\mu = 100$, $\sigma=15$, we have $z=\frac{120 - 100}{15}=\frac{20}{15}\approx1.33$.

Step5: Find proportion for $z = 1.33$

We want $P(X>120)$, which is equivalent to $P(Z > 1.33)$. Since $P(Z>z)=1 - P(Z\leq z)$, looking up $P(Z\leq1.33)$ in the standard - normal table, we find $P(Z\leq1.33)\approx0.9082$. So $P(Z > 1.33)=1 - 0.9082 = 0.0918$.

Answer:

Genius or near genius: $z\approx2.67$; $p = 0.0038$
Very superior intelligence: $z\approx1.33$; $p = 0.0918$