Assume that 150 scores are normally distributed with a mean of 92 and a standards deviation of 11.5

Find the z score corresponding to a score of 115. Show your work.

To find the z-score corresponding to a score of 115, we can use the formula:

z = (x - μ) / σ

Where:
x is the given score (115)
μ is the mean of the distribution (92)
σ is the standard deviation of the distribution (11.5)

Substituting the given values into the formula:

z = (115 - 92) / 11.5

z = 23 / 11.5

z = 2

Therefore, the z-score corresponding to a score of 115 is 2.