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 90 and 120.

Answer 1

Explanation:

Let X be the IQ scores of adults

Given that X is normally distributed with a mean of mu equals 105 and a standard deviation sigma equals 15.

Thus Z score corresponding to any X would be

`(x-105)/(15)`\

Required probability

=probability that a randomly selected adult has an IQ between 90 and 120.

=P(90<X<120)

= P(-1<z<1)

= 0.6836

Answer 2

Answer: 0.6827

Explanation:

Given : Mean IQ score :
`\mu=105`

Standard deviation :
`\sigma=15`

We assume that adults have IQ scores that are normally distributed .

Let x be the random variable that represents the IQ score of adults .

z-score :
`z=(x-\mu)/(\sigma)`

For x= 90

`z=(90-105)/(15)\approx-1`

For x= 120

`z=(120-105)/(15)\approx1`

By using the standard normal distribution table , we have

The p-value :
`P(90<x<120)=P(-1<z<1)`

`P(z<1)-P(z<-1)= 0.8413447-0.1586553=0.6826894\approx0.6827`

Hence, the probability that a randomly selected adult has an IQ between 90 and 120 =0.6827