Question

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose an individual is randomly chosen. a) Find the probability that the person has an IQ greater than 125. b) Find the probability that the person has an IQ score between 105 and 118. c) What is the IQ score of a person whose percentile rank is at the 75th percentile, ?75? d) Use the information from part (c) to fill in the blanks and circle the correct choice in the following statement. ________% of the individuals (persons) have IQ score less than/more than __________ e) “MENSA” is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the “MENSA” organization.

Answer 1

Answer)

As the population is normally distributed we can use standard normal z table to estimate the probability

Given mean = 100

S.d = 15

A)

P(x>125)

First we need to estimate the z score

Z = (x-mean)/s.d

Z = (125-100)/15

Z = 1.67

From z table, P(z<1.67) = 0.9525

As we know sum of all the probabilities is equal to 1

Therefore, P(z>1.67) = 1 - 0.9525 = 0.0475

B)

P(105<x<118)

= p(x<118) - p(x<105)

P(x<118)

Z = (118-100)/15

Z = 1.2

From z table, p(z<1.2) = 0.8849

P(x<105)

Z = (105-100)/15 = 0.33

From z table, p(z<0.33) = 0.6293

Required probability = 0.8849 - 0.6293

= 0.2556

C)

To find the 75 percentile, we need to find the z score which corresponds to 0.75

From z table, p(z<0.67) = 0.67

Therefore, z = 0.67

Z = (x-mean)/s.d

We need to find x here

X = 0.67*15 + 100

X = 110.05

D)

75% iq scores are less than 110.05

E)

To find the top 2% we need to find the z score which corresponds to 98%

That is p(Z<z) = 0.98

From z table, p(z<2.05) = 0.98

Z = (x-mean)/s.d

2.05 = (x-100)/15

X = 130.75

So below 130.75 there is 98% and above 130.75 there is 2%