Question

A test of life satisfaction has a mean of u=60. Jenna's test score of X=72 has a z-score of +3.0. what is the standard deviation for this distribution of test scores? a 6 b 2 c 3 d 4

Answer 1

The standard deviation for the distribution of test scores, where the mean is 60 and Jenna's score of 72 has a z-score of +3.0, is determined to be 4.

Step-by-step explanation:

To determine the standard deviation for the distribution of test scores when the mean is μ=60 and Jenna's test score of X=72 has a z-score of +3.0, we use the formula for the z-score: z = (X - μ) / σ, where X is the individual score, μ is the mean, and σ is the standard deviation.

Plugging in the values we have: 3.0 = (72 - 60) / σ. To solve for σ, we multiply both sides by σ and then divide both sides by 3.0, giving us: σ = (72 - 60) / 3.0 = 12 / 3.0 = 4.

Therefore, the standard deviation for this distribution of test scores is 4.