Question

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100μ=100 and a standard deviation sigma equals 20σ=20. Find the probability that a randomly selected adult has an IQ between 8686 and 114114.

Answer 1

Answer:

51.60% probability that a randomly selected adult has an IQ between 86 and 114.

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 problem, we have that:

$\mu = 114, \sigma = 86$

Find the probability that a randomly selected adult has an IQ between 86 and 114.

Pvalue of Z when X = 114 subtracted by the pvalue of Z when X = 86. So

X = 114

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

$Z = \frac{114 - 100}{20}$

$Z = 0.7$

$Z = 0.7$ has a pvalue of 0.7580

X = 86

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

$Z = \frac{86 - 100}{20}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420

0.7580 - 0.2420 = 0.5160

51.60% probability that a randomly selected adult has an IQ between 86 and 114.