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 (to whole number)?
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 (to whole number)?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**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 (to whole number)?
**Explanation:**
To solve this problem, we need to determine what IQ score corresponds to the top 2% of the distribution. We have:
- Mean (μ) = 100
- Standard Deviation (σ) = 15
For a normal distribution, the Z-score formula is:
\[ Z = \frac{(X - μ)}{σ} \]
Where \( X \) is the score we are trying to find.
For the top 2%, we look for the Z-score that corresponds to the 98th percentile (as 100% - 2% = 98%).
Using a Z-table or calculator, you find that the Z-score corresponding to the 98th percentile is approximately 2.05.
Now, use the Z-score formula to solve for \( X \):
\[ 2.05 = \frac{(X - 100)}{15} \]
Solving for \( X \):
\[ X - 100 = 2.05 \times 15 \]
\[ X - 100 = 30.75 \]
\[ X = 130.75 \]
Therefore, a person must score approximately 131 to qualify for Mensa.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb5aba3c-4931-42d1-80b4-3d9f41ab8cf8%2F462787da-a1ec-4874-8ecf-02450fa8ef45%2Flf0enh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**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 (to whole number)?
**Explanation:**
To solve this problem, we need to determine what IQ score corresponds to the top 2% of the distribution. We have:
- Mean (μ) = 100
- Standard Deviation (σ) = 15
For a normal distribution, the Z-score formula is:
\[ Z = \frac{(X - μ)}{σ} \]
Where \( X \) is the score we are trying to find.
For the top 2%, we look for the Z-score that corresponds to the 98th percentile (as 100% - 2% = 98%).
Using a Z-table or calculator, you find that the Z-score corresponding to the 98th percentile is approximately 2.05.
Now, use the Z-score formula to solve for \( X \):
\[ 2.05 = \frac{(X - 100)}{15} \]
Solving for \( X \):
\[ X - 100 = 2.05 \times 15 \]
\[ X - 100 = 30.75 \]
\[ X = 130.75 \]
Therefore, a person must score approximately 131 to qualify for Mensa.
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