what is the percentage of scores between two numbers if you have a mean and standard deviation

Question

the question is "If the mean is 95 and the standard deviation is 12.5, what percent of employees would have IQs between 90 and 110?

Answer 1

This is a normal distribution with μ = 95 and σ = 12.5.

We need to find two different z-scores using the formula z = (X - μ) / σ.

z₁ = (X₁ - μ) / σ = (90 - 95) / 12.5 = -5/12.5 = -0.4

z₂ = (X₂ - μ) / σ = (110 - 95) / 12.5 = 15/12.5 = 1.2

P(z₁ < -0.4) = 0.3446 and P(z₂ < 1.2) = 0.8849

Subtract the two probabilities and you will get the following.

P(-0.4 < z < 1.2) = 0.8849 - 0.3446 = 0.5403

About 54.03% of the employees will have IQs between 90 and 110.