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?

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).

These z-scores also follow a normal distribution.

The mean is:

The standard deviation is:

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

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2