Question

for a normal distribution with a mean of $mu = 85$ and a standard deviation of $sigma = 20$, find the proportion of the population corresponding to each of the following. a. scores greater than 89

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 89$, $\mu=85$ and $\sigma = 20$.
$z=\frac{89 - 85}{20}=\frac{4}{20}=0.2$

Step2: Find the proportion in the z - table

The z - table gives the proportion of the population to the left of a given z - score. Looking up $z = 0.2$ in the standard normal table, we find that the proportion to the left of $z=0.2$ is $P(Z<0.2)=0.5793$.

Step3: Find the proportion of scores greater than 89

We want $P(X > 89)$, which is equivalent to $P(Z>0.2)$. Since the total area under the normal curve is 1, $P(Z > 0.2)=1 - P(Z<0.2)$.
$P(Z>0.2)=1 - 0.5793 = 0.4207$

Answer:

$0.4207$