Answer:
(a) The distribution of <em>X</em> is <em>N</em> (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 <em>X</em> is defined as the IQ of an individual.
(a)
The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 100 and standard deviation, <em>σ</em> = 15.
The probability density function of <em>X</em> is:
Thus, the distribution of <em>X</em> is <em>N</em> (100, 15²).
(b)
Compute the probability that a person has an IQ greater than 130 as follows:
Thus, the probability that a person has an IQ greater than 130 is 0.0228.
(c)
Let <em>x</em> 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 <em>z</em> is:
<em>z</em> = 2.06.
Compute the value of <em>x</em> as follows:
Thus, the minimum IQ needed to qualify for the Mensa organization is 131.
(d)
Let <em>x</em>₁ and <em>x</em>₂ be the values between which middle 20% of IQs fall.
This implies that:
The value of <em>z</em> is:
<em>z</em> = 0.26.
Compute the value of <em>x</em> as follows:
Thus, the middle 20% of IQs fall between 96 and 104.