Question

8. Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Let X be the IQ scores for adults. So ( ) 2 X N~ 100,15

a. Find the probability that a randomly selected adult has an IQ greater than 131.5 (the requirement for membership in the Mensa organization). Show the area under the normal curve.

b. Find the probability that a randomly selected adult has an IQ between 90 and 110 (referred to as the normal range). Show the area under the normal curve.

c. What is the IQ score separating the bottom 20% from the top 80%. Use the normal curve to explain your answer.

Answer 1

solution:

Given data

Mean () = 100

Standard deviation () = 15

Let X be the normal random variable representing IQ scores of adults

a) Probability that a randomly selected adult has an IQ greater than 131.5 = P(X>131.5)

=

= P(Z>2.1)

= 1 - P(Z<=2.1)

= 1 - 0.9821 [using standard normal disribution table ]

= 0.0179

Probability that a randomly selected adult has an IQ greater than 131.5 = 0.0179

b) Probability that a randomly selected adult has an IQ score between 90 and 110 = P(90<X<110)

= P(X<110) - P(X<90)

=  

= P(Z<0.6667) - P(Z<-0.6667)

= P(Z<0.6667) - P(Z>0.6667)

= P(Z<0.6667) - [ 1 - P(Z<=0.6667) ]

=2 P(Z<0.6667) - 1

= 2*0.7475 - 1 [using standard normal disribution table ]

= 0.495

Probability that a randomly selected adult has an IQ score between 90 and 110 = 0.495

c) Let X1 be the IQ score separating the bottom 20% from top 80%

then P(X<X1) = 20% = 0.20

   = 0.20

= - 0.8415    [using standard normal disribution table ]

X1 - 100 = -12.6225

X1 = 87.3775

Here 87.3775 is an IQ score that is separating the bottom 20% from the top 80% i.e., 20% IQ are below this point and 80% of IQs are above this IQ.