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1) IQ is normally distributed with a mean of 100 and a standard deviation of 15.

ID: 2922036 • Letter: 1

Question

1) 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) ive the distribution of X.

b) Find the probability that the person has an IQ greater than 140.

Write the probability statement. P (___)

What is the probability? (Round your answer to 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 =

d) The middle 30% 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 =

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 100
standard Deviation ( sd )= 15
a.
P(X > 140) = (140-100)/15
= 40/15 = 2.6667
= P ( Z >2.6667) From Standard Normal Table
= 0.0038
c.
P ( Z > x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is 2.0537
P( x-u / (s.d) > x - 100/15) = 0.02
That is, ( x - 100/15) = 2.0537
--> x = 2.0537 * 15+100 = 130.8062
d.
P ( Z < x ) = 0.4
Value of z to the cumulative probability of 0.4 from normal table is -0.2533
P( x-u/s.d < x - 100/15 ) = 0.4
That is, ( x - 100/15 ) = -0.2533
--> x = -0.2533 * 15 + 100 = 96.1998
GREATER THAN
P ( Z > x ) = 0.4
Value of z to the cumulative probability of 0.4 from normal table is 0.2533
P( x-u / (s.d) > x - 100/15) = 0.4
That is, ( x - 100/15) = 0.2533
--> x = 0.2533 * 15+100 = 103.8002

P(x1 < X < x2) = P( 96.1998 < x < 103.8002) ~ P( 96 < x < 104)