Question

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100 and a standard deviation sigma equals 20. Find the probability that a randomly selected adult has an IQ less than 132. The probability that a randomly selected adult has an IQ less than 132 is?

Answer 1

Step-by-step explanation:

First find the z-score.

z = (x - μ) / σ

z = (132 - 100) / 20

z = 1.6

From a z-score table:

P(z<1.6) = 0.9452

So there's a 94.52% probability that a randomly selected adult will have an IQ less than 132.

Answer 2

Answer:

There is a 94.52% probability that a randomly selected adult has an IQ less than 132.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 20. This means that $\mu = 100, \sigma = 20$.

The probability that a randomly selected adult has an IQ less than 132 is?

This probability is the pvalue of Z when $X = 132$. So:

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

$Z = \frac{132 - 100}{20}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452.

This means that there is a 94.52% probability that a randomly selected adult has an IQ less than 132.