Question

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equals 15. Find the probability that a randomly selected adult has an IQ between 87 and 123.

Answer 1

Answer:

0.7698

Step-by-step explanation:

If you call your random variable $X$, then what you are looking for is

$P(87 \leq X \leq 123)$

because you want the probability of $X$ being between 87 and 123.

We need a table with of the normal distribution. But we can only find the table with $\mu = 0$ and $\sigma = 1$. Because of that, first we need to  normalize our random variable:

$Z = \frac{X - \mu}{\sigma} = \frac{X - 105}{15}$

(you can always normalize your variable following the same formula!)

now we can do something similar to our limits, to get a better expression:

$\frac{87 - 105}{15} = \frac{-18}{15} = -1.2$

$\frac{123 - 105}{15} = \frac{18}{15} = 1.2$

And we transform our problem to a simpler one:

$P(87 \leq X \leq 123) = P(-1.2 \leq Z \leq 1.2) = P(Z \leq 1.2) - P(Z \leq -1.2)$

(see Figure 1)

From our table we can see that $P(Z \leq 1.2) = 0.8849$ (this is represented in figure 2).

Remember that the whole area below the curve is exactly 1. So we can conclude that  $P(Z \geq 1.2) = 0.1151$ (because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then

$P(Z \leq -1.2)= P(Z \geq 1.2) = 0.1151$.

FINALLY:

$P(Z \leq 1.2) - P(Z \leq -1.2) = 0.8849 - 0.1151 = 0.7698$