A person must score in the upper of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. There are Mensa members in countries throughout the world (Mensa International website, January , ). If IQ scores are normally distributed with a mean of and a standard deviation of , what score must a person have to qualify for Mensa (to whole number)?

The person must have a score of 131 to be able to qualify for Mensa

Step-by-step explanation:

The complete and correct question is as follows;

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. There are 110,000 Mensa members in 100 countries throughout the world (Mensa International website, January 8, 2013). 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(to whole number)

Solution

Firstly, we calculate the z score from the given left tailed area. Since our upper limit is 2%, left tailed area will be 98% which is simply 0.98

Left tailed area = 0.98

Then, using standard score table,

z = 2.054

Mathematically; x = u + z * s

According to this question;

u = mean = 100

z = the critical z score = 2.054

s = standard deviation = 15

Substituting these values into the equation, we have

x = 100 + 15(2.054)

x = 100 + 30.81 = 130.81

This is 131 to whole number

is only a line? i think so

eve and marks ratios are equivalent

step-by-step explanation: