Answer: 0.6827
Step-by-step explanation:
Given : Mean IQ score : [tex]\mu=105 [/tex]
Standard deviation : [tex]\sigma=15[/tex]
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 : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 90
[tex]z=\dfrac{90-105}{15}\approx-1[/tex]
For x= 120
[tex]z=\dfrac{120-105}{15}\approx1[/tex]
By using the standard normal distribution table , we have
The p-value : [tex]P(90<x<120)=P(-1<z<1)[/tex]
[tex]P(z<1)-P(z<-1)= 0.8413447-0.1586553=0.6826894\approx0.6827[/tex]
Hence, the probability that a randomly selected adult has an IQ between 90 and 120 =0.6827
Answer:
Step-by-step 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
[tex]\frac{x-105}{15}[/tex]\
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
