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.