Question

Assume that adults have IQ scores that are normally distributed with a mean of 95.695.6 and a standard deviation of 19.519.5. Find the probability that a randomly selected adult has an IQ greater than 115.9115.9

Answer 1

Answer:

0.14917

Step-by-step explanation:

We have been given that adults have IQ scores that are normally distributed with a mean of 95.6 and a standard deviation of 19.5. We are asked to find the probability that a randomly selected adult has an IQ greater than 115.9.

First of all, we will find z-score corresponding to 115.9 using z-score formula.

$z=\frac{x-\mu}{\sigma}$, where

z = z-score,

x = Random sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

Upon substituting our given values in z-score formula, we will get:

$z=\frac{115.9-95.6}{19.5}$

$z=\frac{20.3}{19.5}$

$z=1.04$

Now, we will use normal distribution table to find the $P(z>1.04)$.

Using formula $P(z>a)=1-P(z$, we will get:

$P(z>1.04)=1-P(z$

$P(z>1.04)=1-0.85083$

$P(z>1.04)=0.14917$

Therefore, the probability that a randomly selected adult has an IQ greater than 115.9 is 0.14917 or approximately 14.92%.