Question

1. 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 b. scores less than 72 c. scores between 70 and 100

Explanation:

Step1: Calculate z - score formula

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

Step2: Solve part a

For $x = 89$, $\mu=85$ and $\sigma = 20$. Calculate the z - score:
$z=\frac{89 - 85}{20}=\frac{4}{20}=0.2$
We want $P(X>89)$, which is equivalent to $P(Z > 0.2)$. Using the standard normal table, $P(Z>0.2)=1 - P(Z\leqslant0.2)$. From the standard - normal table, $P(Z\leqslant0.2)=0.5793$. So $P(Z > 0.2)=1 - 0.5793 = 0.4207$.

Step3: Solve part b

For $x = 72$, $\mu = 85$ and $\sigma=20$. Calculate the z - score:
$z=\frac{72 - 85}{20}=\frac{-13}{20}=-0.65$
We want $P(X < 72)$, which is equivalent to $P(Z<-0.65)$. From the standard - normal table, $P(Z<-0.65)=0.2578$.

Step4: Solve part c

For $x_1 = 70$, $z_1=\frac{70 - 85}{20}=\frac{-15}{20}=-0.75$
For $x_2 = 100$, $z_2=\frac{100 - 85}{20}=\frac{15}{20}=0.75$
We want $P(70$P(-0.75

Answer:

a. $0.4207$
b. $0.2578$
c. $0.5468$