Question

A distribution of scores for a test of life stressors has a mean of µ = 125 and standard deviation of σ = 15. The researcher calculates z-scores to standardize the distribution. What are the mean and the standard deviation for the z-scores in this distribution?

Answer 1

Answer:

The mean and standard deviation for the z-scores in this distribution are 0 and 1 respectively.

Step-by-step explanation:

Let the random variable X follow a Normal distribution with mean μ and standard deviation σ.

The z-scores are standardized form of the raw scores X. It is computed by subtracting the mean (μ) from the raw score x and dividing the result by the standard deviation (σ).

$z=\frac{x-\mu}{\sigma}$

These z-scores also follow a normal distribution.

The mean is:

$E(z)=E[\frac{x-\mu}{\sigma} ]=\frac{1}{\sigma}\times [E(x)-\mu] =\frac{1}{\sigma}\times [\mu-\mu]=0$

The standard deviation is:

$Var(z)=Var[\frac{x-\mu}{\sigma} ]=\frac{1}{\sigma^{2}}\times [Var(x)-Var(\mu)] =\frac{\sigma^{2}-0}{\sigma^{2}}=1\\SD(z)=\sqrt{Var(z)}=1$

Thus, the mean and standard deviation for the z-scores in this distribution are 0 and 1 respectively.