Question

Suppose IQ scores are normally distributed with a mean value of 100 and a standard deviation of 12.

Use the Empirical Rule:

i. 68% of people have an IQ between 88 and 112.
ii. 95% of people have an IQ between 76 and 124.
iii. 99.7% of people have an IQ between 64 and 136.

What percent of individuals have an IQ above 125?

Answer 1

Approximately 1.9% of individuals have an IQ above 125.

Step-by-step explanation:

To find the percent of individuals with an IQ above 125, we first need to find the z-score for 125. The formula for calculating the z-score is: z = (x - μ) / σ, where x is the IQ score, μ is the mean, and σ is the standard deviation. In this case, x = 125, μ = 100, and σ = 12.

Using the formula, we can calculate the z-score as follows:

z = (125 - 100) / 12 = 25 / 12 ≈ 2.08

Next, we can use a standard normal distribution table or calculator to find the area to the right of the z-score. The area to the right represents the percentage of individuals with an IQ score above 125. Using a standard normal distribution table, we find that the area to the right of 2.08 is approximately 0.019. This means that approximately 1.9% of individuals have an IQ above 125.