Assume that adults have iq scores that are normally distributed with a mean of 105 and a standard deviation of 20. Find the probability that a randomly selected adult has an iq between 89 and 121

Answer 1

Since we assume these data are normally distributed in the adult population, you can answer this question by first converting your given scores to a standard normal distribution (that is, a set of z-scores):

x₁ = 89,

x₂ = 121,

mean(z) = 105,

sigma = 20.

z₁ = ( x₁ - mean(x) ) / sigma = (89 - 105) / 20 = -0.80, and

z₂ = ( x₂ - mean(x) ) / sigma = (121 - 105) / 20 = 0.80.

Next, we would like to find the area of standard normal distribution that lies between -0.8 and +0.8.

There are several ways to do this using z-tables in the back of your textbook (or R software). Using a z-table you often have choices between the "body" of the distribution (either the upper- or lower-tail), in which case will come to the same conclusion using subtraction. However, most textbooks also have a section on "the area between mean and z". This is what we will focus on.

Looking at the back of your textbook for the z-table, notice that the area between the mean and z for a z-score of 0.80 = .2881.

You may also notice that your table does not have negative z-scores. Why? Because the positive half of the standard normal distribution is a mirror image of itself; therefore, every z-score with a value less than zero will correspond in analogous values in the positive direction. So to answer your question, you simply add the two areas together:

area for z₁ + area for z₂ = 0.2881 + 0.2881 = 0.5762.