Question

1. a population forms a normal distribution with a mean of $mu = 85$ and a standard deviation of $sigma = 24$. for each of the following samples, compute the $z$-score for the sample mean.

a. $m = 91$ for $n = 4$ scores
b. $m = 91$ for $n = 9$ scores
c. $m = 91$ for $n = 16$ scores
d. $m = 91$ for $n = 36$ scores

Explanation:

Step1: Calculate standard error

The formula for standard error $\sigma_M=\frac{\sigma}{\sqrt{n}}$.
For part a: $\sigma = 24$, $n = 4$, so $\sigma_M=\frac{24}{\sqrt{4}}=\frac{24}{2}=12$.
For part b: $\sigma = 24$, $n = 9$, so $\sigma_M=\frac{24}{\sqrt{9}}=\frac{24}{3}=8$.
For part c: $\sigma = 24$, $n = 16$, so $\sigma_M=\frac{24}{\sqrt{16}}=\frac{24}{4}=6$.
For part d: $\sigma = 24$, $n = 36$, so $\sigma_M=\frac{24}{\sqrt{36}}=\frac{24}{6}=4$.

Step2: Calculate z - score

The formula for z - score $z=\frac{M-\mu}{\sigma_M}$. Given $\mu = 85$, $M = 91$.
For part a: $z=\frac{91 - 85}{12}=\frac{6}{12}=0.5$.
For part b: $z=\frac{91 - 85}{8}=\frac{6}{8}=0.75$.
For part c: $z=\frac{91 - 85}{6}=\frac{6}{6}=1$.
For part d: $z=\frac{91 - 85}{4}=\frac{6}{4}=1.5$.

Answer:

a. $z = 0.5$
b. $z = 0.75$
c. $z = 1$
d. $z = 1.5$