Question

In a recently administered iq test, the scores were distributed normally, with mean 100 and standard deviation 15. what proportion of the test takers scored between 70 and 130?

Answer 1

To solve for proportion we make use of the z statistic. The procedure is to solve for the value of the z score and then locate for the proportion using the standard distribution tables. The formula for z score is:

z = (X – μ) / σ

where X is the sample value, μ is the mean value and σ is the standard deviation

 

when X = 70

z1 = (70 – 100) / 15 = -2

Using the standard distribution tables, proportion is P1 = 0.0228

 

when X = 130

z2 = (130 – 100) /15 = 2

Using the standard distribution tables, proportion is P2 = 0.9772

 

Therefore the proportion between X of 70 and 130 is:

P (70<X<130) = P2 – P1

P (70<X<130) = 0.9772 - 0.0228

P (70<X<130) = 0.9544

 

Therefore 0.9544 or 95.44% of the test takers scored between 70 and 130.