Question

On a test that has a normal distribution, a score of 19 falls one standard deviation below the mean, and a score of 40 falls two standard deviations above the mean. Determine the mean of this test.

Answer 1

Hi there!

We can use the following equation:

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

z = amount of standard deviations away a value is from the mean (z-score)

σ = standard deviation

x = value

μ = mean

Plug in the knowns for both and rearrange to solve for the mean:

$-1 = \frac{19-\mu}{\sigma}\\\\-\sigma = 19 - \mu\\\\ \sigma = -19 + \mu$

Other given:

$2 = \frac{40-\mu}{\sigma}\\\\2\sigma = 40- \mu\\\\ \sigma = 20 - \frac{\mu}{2}$

Set both equal to each other and solve:

$-19 + \mu = 20 - \frac{\mu}{2}\\\\\frac{3\mu}{2} = 39 \\\\3\mu = 78 \\\\\boxed{\mu = 26 }$