Question

Assume that adults have IQ scores that are normally distributed with a mean of 96 and a standard deviation of 15.7. Find the probability that randomly selected adults has and IQ greater than 123.4

Answer 1

Answer:

4.05% probability that a randomly selected adult has an IQ greater than 123.4.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 96, \sigma = 15.7$

Probability that a randomly selected adult has an IQ greater than 123.4.

This is 1 subtracted by the pvalue of Z when X = 123.4. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{123.4 - 96}{15.7}$

$Z = 1.745$

$Z = 1.745$ has a pvalue of 0.9595

1 - 0.9595 = 0.0405

4.05% probability that a randomly selected adult has an IQ greater than 123.4.