Complete Question
Assume that adults have IQ scores that are normally distributed with a mean μ=100 and a standard deviation σ=15. Find the probability that a randomly selected adult has an IQ between 81 and 119.
Answer:
The probability is [tex]P( x_1 < X < x_2) = 0.79474[/tex]
Step-by-step explanation:
From the question we are told that
The standard deviation is σ = 15.
The mean μ= 100
The range we are considering is [tex]x_1 = 81 , \ x_2 = 119[/tex]
Now given that IQ scores are normally distributed
Then the probability that a randomly selected adult has an IQ between 81 and 119 is mathematically represented as
[tex]P( x_1 < X < x_2) = P(\frac{x_1 - \mu }{\sigma } <\frac{X - \mu }{\sigma } < \frac{x_2- \mu }{\sigma } )[/tex]
Generally
[tex]\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of \ X )[/tex]
So
[tex]P( x_1 < X < x_2) = P(\frac{x_1 - \mu }{\sigma } <Z < \frac{x_2- \mu }{\sigma } )[/tex]
substituting values
[tex]P( x_1 < X < x_2) = P(\frac{81 - 100 }{15 } <Z < \frac{119- 100 }{15 } )[/tex]
[tex]P( x_1 < X < x_2) = P( -1.2667 <Z <1.2667 )[/tex]
[tex]P( x_1 < X < x_2) = P(Z <1.2667 )-P( Z < -1.2667 )[/tex]
From the standardized Z table
[tex]P(Z <-1.2667 ) = 0.10263[/tex]
And [tex]P(Z <1.2667 ) = 0.89737[/tex]
So
[tex]P( x_1 < X < x_2) = 0.89737 - 0.10263[/tex]
[tex]P( x_1 < X < x_2) = 0.79474[/tex]
