Question

A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa?

Answer 1

We have given the normal distribution

μ = mean = 100

σ = standard deviation = 15

We have given the top area as 2 %

We convert the top area in to decimal = 0.02

We find the bottom area

We know total area is 1

Bottom area = 1- top area

Bottom area = 1- 0.02 = 0.9800

We use Z table to find the z score for the area 0.9800

We look for the area which is close to 0.9800

Area 0.9798 is very close to 0.9800

Z score = row headed number to 0.9798 + column headed number to 0.9798

Z score = 2.0 + 0.05

Z score = 2.05

Now we use the formula of the z score

Final answer :-

What score must a person have to qualify for Mensa = 130.75

Note :- Round your answer according to instructions

If they have mention round to whole number then answer would be 131

I hope this will help you :)