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?

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.

10010010001010010101001001010010100101010010101010010010101010110010001011010010010001

step-by-step explanation:

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead