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: $\sigma = 12$, $n = 36$, so $\sigma_M=\frac{12}{\sqrt{36}}=\frac{12}{6}=2$.
For part b: $\sigma = 30$, $n = 36$, so $\sigma_M=\frac{30}{\sqrt{36}}=\frac{30}{6}=5$.

Step2: Calculate the z - score

The formula for the z - score of a sample mean is $z=\frac{M-\mu}{\sigma_M}$.
Given $\mu = 65$, $M = 59$.
For part a: $z=\frac{59 - 65}{2}=\frac{-6}{2}=-3$.
For part b: $z=\frac{59 - 65}{5}=\frac{-6}{5}=-1.2$.

Step3: Determine if the sample mean is typical or extreme

Typically, values with $|z|\leq2$ are considered typical (representative), and values with $|z|>2$ are considered extreme.
For part a: $|z| = |-3|=3>2$, so the sample mean is an extreme value.
For part b: $|z|=|-1.2| = 1.2\leq2$, so the sample mean is a typical (representative) 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 (representative) value.