Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Mensa is an international society that has one - and only one - qualification for membership: a score in the top 2% of the population on an IQ test. (a) What IQ score should one have in order to be eligible for Mensa? (b) In a typical region of 145,000 people, how many are eligible for Mensa

Answer 1

To be eligible for Mensa, one must have an IQ score in the top 2% of the population. With a mean IQ score of 100 and a standard deviation of 15, the IQ score needed for Mensa is approximately 131. In a typical region of 145,000 people, approximately 2,900 would be eligible for Mensa.

Step-by-step explanation:

To be eligible for Mensa, one must have an IQ score in the top 2% of the population. With a mean IQ score of 100 and a standard deviation of 15, we can use the formula: IQ score = mean + (z-score * standard deviation). To find the IQ score needed for Mensa, we need to find the z-score corresponding to the top 2% of the population. Using a z-table, we find that the z-score is approximately 2.05. Plugging this value into the formula, we have:

IQ score = 100 + (2.05 * 15) = 130.75

Therefore, one should have an IQ score of approximately 131 in order to be eligible for Mensa.

To find the number of people eligible for Mensa in a typical region of 145,000 people, we can use the formula: number of eligible people = population size * proportion of population with IQ scores above Mensa threshold. Since the top 2% of the population is eligible for Mensa, the proportion of population with IQ scores above the Mensa threshold is 0.02. Plugging in the values, we have:

Number of eligible people = 145,000 * 0.02 = 2,900

Therefore, in a typical region of 145,000 people, approximately 2,900 would be eligible for Mensa.

Answer 2

a) In order to be eligible for Mensa, a person must have an IQ score of at least 130.81.

b) 2900 people are elegible for Mensa.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 100, \sigma = 15`

(a) What IQ score should one have in order to be eligible for Mensa?

Scores in at least the 98th percentile, so scores of at least X when Z has a pvalue of 0.98. So at least X when Z = 2.054.

`Z = (X - \mu)/(\sigma)`

`2.054 = (X - 100)/(15)`

`X - 100 = 2.054*15`

`X = 130.81`

In order to be eligible for Mensa, a person must have an IQ score of at least 130.81.

(b) In a typical region of 145,000 people, how many are eligible for Mensa

Only 2% are elegible. So

0.02*145000 = 2900

2900 people are elegible for Mensa.