Respuesta :
Answer:
(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.
Step-by-step 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:
[tex]f_{X}(x)=\frac{1}{15\sqrt{2\pi}}\times e^{-(x-100)^{2}/(2\times 15^{2})};\ -\infty<X<\infty[/tex]
Thus, the distribution of X is N (100, 15²).
(b)
Compute the probability that a person has an IQ greater than 130 as follows:
[tex]P(X>130)=P(\frac{X-\mu}{\sigma}>\frac{130-100}{15})[/tex]
[tex]=P(Z>2)\\=1-P(Z<2)\\=1-0.97725\\=0.02275\\\approx 0.0228[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma}\\2.06=\frac{x-100}{15}\\x=100+(2.06\times 15)\\x=130.9\\x\approx131[/tex]
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:
[tex]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[/tex]
The value of z is:
z = 0.26.
Compute the value of x as follows:
[tex]-z=\frac{x_{1}-\mu}{\sigma}\\-0.26=\frac{x_{1}-100}{15}\\x_{1}=100-(0.26\times 15)\\x=96.1\\x\approx96[/tex] [tex]z=\frac{x_{2}-\mu}{\sigma}\\0.26=\frac{x_{2}-100}{15}\\x_{2}=100+(0.26\times 15)\\x=103.9\\x\approx104[/tex]
Thus, the middle 20% of IQs fall between 96 and 104.
