Question

1. a sample of ( n = 36 ) scores is selected from a normal distribution with a mean of ( mu = 65 ). compute the ( z )-score for a sample mean of ( m = 59 ) and determine whether the sample mean is a typical, representative value or an extreme value for each of the following: a. a population standard deviation of ( sigma = 12 ) b. a population standard deviation of ( sigma = 30 )

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_M=\frac{\sigma}{\sqrt{n}}$.

For part a:

Given $\sigma = 12$ and $n = 36$, then $\sigma_M=\frac{12}{\sqrt{36}}=\frac{12}{6}=2$.

For part b:

Given $\sigma = 30$ and $n = 36$, then $\sigma_M=\frac{30}{\sqrt{36}}=\frac{30}{6}=5$.

Step2: Calculate the z - score

The formula for the z - score of a sample mean is $z=\frac{M-\mu}{\sigma_M}$.
Given $\mu = 65$ and $M = 59$.

For part a:

Substitute $\mu = 65$, $M = 59$ and $\sigma_M=2$ into the formula: $z=\frac{59 - 65}{2}=\frac{-6}{2}=-3$.

For part b:

Substitute $\mu = 65$, $M = 59$ and $\sigma_M = 5$ into the formula: $z=\frac{59 - 65}{5}=\frac{-6}{5}=-1.2$.

Step3: Determine if the value is typical or extreme

Typically, values with $|z|\leq2$ are considered typical (within 2 standard errors of the mean), and values with $|z|>2$ are considered extreme.

For part a:

Since $|z| = |-3|=3>2$, the sample mean is an extreme value.

For part b:

Since $|z|=|-1.2| = 1.2\leq2$, the sample mean is a typical value.

Answer:

a. The z - score is $z=-3$, and the sample mean is an extreme value.
b. The z - score is $z = -1.2$, and the sample mean is a typical value.