Question

Mensa is a high IQ society that admits people as members if they can score at the 98th percentile or above on certain standardized IQ (intelligence quotient) tests. On one such test, the Stanford Binet, the qualifying score is 132. The test consists of n questions, each with m choices.

a) On any given test question, the person taking the test knows the answer with probability p. Assume that when the person does not know the answer, the person guesses an option completely at random. Calculate the probability a person knew the answer to a question, given that they answered it correctly.

b) If a person receives a score of 132 or higher on the test, they are considered to have an IQ of 132 or higher. However, individuals with IQ less than 132 can also receive such scores about 0.1% of the time due to lucky guessing. Given that a person is labeled as having IQ of 132 or higher, what is the probability they actually have IQ below 132? Assume that all individuals with IQ of 132 or higher receive an accurate score 95% of the time.

Answer 1

Explanation:

a)

Let C represent a question answered correctly

G represent answer is given by guessing

K represent that individual is knowing the answer

Now given that

P(K)=p

So P(G)=1-p

Also if an individual know the answer then he will give the correct answer correctly hence P(C|K)=1

While if one is guessing then he will give correct option with 1/m probability that there are m options hence

P(C|G)=1/m

We have to find the P(K|C)

Now using Bayes theorem

P(K/C) = P(K)×P(C/K) = p×1

P(K)×P(C/K)+P(G)×P(C/G) p×1 +(1-p)×1/m

= mp

1+p(m-1)

b)

Let

A represent that a person IQ level is more than 132

B represent IQ level is less than 132

C represent person lebelled as have IQ more then 132

Any person to be in that society needs to be in 98th percentile hence 98% peoples will be below that level hence

P(A)=0.02 P(B)=0.98

Also given than P(C|A)=0.95 while P(C|B)=0.001

We have to find P(B|C)

Now using Bayes theorem

P(B/C)= P(B)×P(C/B) = 0.98×0.001

P(B)×P(C/B)+P(A)×P(C/A) 0.98×0.001 +0.02×0.95

=0.045