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: Recall the formula for the z - score of the sample mean

The formula for the z - score of the sample mean is \(z=\frac{M-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(M\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step2: Substitute the values for part a

Given \(\mu = 85\), \(\sigma=24\), \(M = 91\), \(n = 4\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{24}{\sqrt{4}}=\frac{24}{2}=12\)
Then, \(z=\frac{91 - 85}{12}=\frac{6}{12}=0.5\)

Step3: Substitute the values for part b

Given \(\mu = 85\), \(\sigma=24\), \(M = 91\), \(n = 9\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{24}{\sqrt{9}}=\frac{24}{3}=8\)
Then, \(z=\frac{91 - 85}{8}=\frac{6}{8}=0.75\)

Step4: Substitute the values for part c

Given \(\mu = 85\), \(\sigma=24\), \(M = 91\), \(n = 16\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{24}{\sqrt{16}}=\frac{24}{4}=6\)
Then, \(z=\frac{91 - 85}{6}=\frac{6}{6}=1\)

Step5: Substitute the values for part d

Given \(\mu = 85\), \(\sigma=24\), \(M = 91\), \(n = 36\)
First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{24}{\sqrt{36}}=\frac{24}{6}=4\)
Then, \(z=\frac{91 - 85}{4}=\frac{6}{4}=1.5\)

Answer:

a. \(z = 0.5\)
b. \(z=0.75\)
c. \(z = 1\)
d. \(z = 1.5\)