Question

Assume adults have IQ scores that are normally distributed with a mean of 102 and standard deviation of 16. Find the probability that a randomly selected individual has an IQ between 81 and 109

Answer 1

Answer:

57.49% probability that a randomly selected individual has an IQ between 81 and 109

Step-by-step explanation:

When the distribution is normal, we use 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:

$\mu = 102, \sigma = 16$

Find the probability that a randomly selected individual has an IQ between 81 and 109

This is the pvalue of Z when X = 109 subtracted by the pvalue of Z when X = 81. So

X = 109

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

$Z = \frac{109 - 102}{16}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

X = 81

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

$Z = \frac{81 - 102}{16}$

$Z = -1.31$

$Z = -1.31$ has a pvalue of 0.0951

0.67 - 0.0951 = 0.5749

57.49% probability that a randomly selected individual has an IQ between 81 and 109