Question

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.

a. Give the distribution of X
b. Find the probability that the person has an IQ greater than 130, write the probability statement. What is the probability? (round your answer to the four decimal places.)
c. Mensa is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the Mensa organization.

Write the probability statement.

P(X > x) =

What is the minimum IQ? (Round your answer to the nearest whole number.)
x =  
Sketch the graph

d. The middle 20% of IQs fall between what two values?

Write the probability statement.
P(x1 < X < x2) =

State the two values. (Round your answers to the nearest whole number.)
x1=x2=
Sketch the graph

Answer 1

(a) The distribution of X is N (100, 15²).

(b) The probability that a person has an IQ greater than 130 is 0.0228.

(c) The minimum IQ needed to qualify for the Mensa organization is 131.

(d) The middle 20% of IQs fall between 96 and 104.

Explanation:

The random variable X is defined as the IQ of an individual.

(a)

The random variable X is normally distributed with mean, μ = 100 and standard deviation, σ = 15.

The probability density function of X is:

`f_(X)(x)=(1)/(15√(2\pi))* e^{-(x-100)^(2)/(2* 15^(2))};\ -\infty<X<\infty`

Thus, the distribution of X is N (100, 15²).

(b)

Compute the probability that a person has an IQ greater than 130 as follows:

`P(X>130)=P((X-\mu)/(\sigma)>(130-100)/(15))`

`=P(Z>2)\\=1-P(Z<2)\\=1-0.97725\\=0.02275\\\approx 0.0228`

Thus, the probability that a person has an IQ greater than 130 is 0.0228.

(c)

Let x represents the top 2% of all IQs.

Then, P (X > x) = 0.02.

⇒ P (X < x) = 1 - 0.02

⇒ P (Z < z) = 0.98

The value of z is:

z = 2.06.

Compute the value of x as follows:

`z=(x-\mu)/(\sigma)\\2.06=(x-100)/(15)\\x=100+(2.06* 15)\\x=130.9\\x\approx131`

Thus, the minimum IQ needed to qualify for the Mensa organization is 131.

(d)

Let x₁ and x₂ be the values between which middle 20% of IQs fall.

This implies that:

`P(x_(1)<X<x_(2))=0.20\\P(-z<Z<z)=0.20\\P(Z<z)-P(Z<-z)=0.20\\P(Z<z)-[1-P(Z<z)]=0.20\\2P(Z<z)=1.20\\P(Z<z)=0.60`

The value of z is:

z = 0.26.

Compute the value of x as follows:

`-z=(x_(1)-\mu)/(\sigma)\\-0.26=(x_(1)-100)/(15)\\x_(1)=100-(0.26* 15)\\x=96.1\\x\approx96`
`z=(x_(2)-\mu)/(\sigma)\\0.26=(x_(2)-100)/(15)\\x_(2)=100+(0.26* 15)\\x=103.9\\x\approx104`

Thus, the middle 20% of IQs fall between 96 and 104.