What is the likelihood of selecting adults with an IQ over 130 in a sample of 200?

To determine the likelihood of selecting adults with an IQ over 130 from a population where the mean IQ is 112 and the standard deviation is 20, we can use the properties of the normal distribution.

The first step is to calculate the z-score for an IQ of 130 using the formula:

z = (X - μ) / σ

Where:

• X is the value we are interested in (130),
• μ is the mean (112),
• σ is the standard deviation (20).

Plugging in the numbers:

z = (130 - 112) / 20 = 0.9

Next, we can look up the z-score of 0.9 in the standard normal distribution table, or use a calculator, which gives us approximately 0.8159. This value represents the probability of selecting an adult with an IQ less than 130.

To find the probability of selecting an adult with an IQ greater than 130, we need to subtract this value from 1:

P(X > 130) = 1 - P(X < 130) = 1 - 0.8159 = 0.1841

This means that about 18.41% of the adult population has an IQ over 130. If we multiply this probability by the sample size of 200, we can estimate the expected number of adults with an IQ over 130:

Expected number = 0.1841 * 200 ≈ 36.82

So, in a random sample of 200 adults, we can expect around 37 individuals to have an IQ above 130.