Question

the distribution of certain test scores is a nonstandard normal distribution with a mean of 50 and a standard deviation of 6. what are the values of the mean and standard deviation after all test scores have been standardized by converting them to z - scores using ( z=\frac{x - mu}{sigma} )?
a. the mean is 100 and the standard deviation is 10.
b. the mean is 1 and the standard deviation is 0.
c. the mean is 10 and the standard deviation is 100.
d. the mean is 0 and the standard deviation is 1.

Explanation:

Step1: Recall the properties of z - scores

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation of the original distribution.

Step2: Calculate the mean of z - scores

Let \(X\) be a random variable with mean \(\mu\) and standard deviation \(\sigma\). The mean of \(Z=\frac{X-\mu}{\sigma}\) is \(E(Z)=E(\frac{X-\mu}{\sigma})=\frac{1}{\sigma}(E(X)-\mu)\). Since \(E(X)=\mu\), then \(E(Z)=\frac{\mu - \mu}{\sigma}=0\).

Step3: Calculate the standard deviation of z - scores

The variance of \(Z\) is \(Var(Z)=Var(\frac{X-\mu}{\sigma})=\frac{1}{\sigma^{2}}Var(X)\). Since \(Var(X)=\sigma^{2}\), then \(Var(Z) = 1\). And the standard deviation of \(Z\) (because standard deviation is the square - root of variance) is \(\sqrt{Var(Z)} = 1\).

Answer:

D. The mean is 0 and the standard deviation is 1.