11. MENSA, the high IQ society, only accepts membership into their organization if an applicant's IQ is at or above the 98th percentile. If IQ scores follow a normal distribution with a mean of 100 and a standard deviation of 15, what score would someone need in order to be eligible for MENSA membership?
11. MENSA, the high IQ society, only accepts membership into their organization if an applicant's IQ is at or above the 98th percentile. If IQ scores follow a normal distribution with a mean of 100 and a standard deviation of 15, what score would someone need in order to be eligible for MENSA membership?
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Problem 1P
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![**Understanding Mensa Membership Requirements**
**Question:**
Mensa, the high IQ society, only accepts membership into their organization if an applicant's IQ is at or above the 98th percentile. If IQ scores follow a normal distribution with a mean of 100 and a standard deviation of 15, what score would someone need in order to be eligible for Mensa membership?
**Explanation:**
To determine eligibility for Mensa based on IQ scores, understanding percentiles and normal distribution is key. The problem specifies that IQ scores are distributed normally, which means they form a bell-shaped curve centered around the mean. Here's a breakdown of the terms:
1. **Mean (μ):** The average score. For IQ, this is 100.
2. **Standard Deviation (σ):** A measure of how spread out the scores are around the mean. For IQ, the standard deviation is 15.
**Finding the Score Required:**
To find the score corresponding to the 98th percentile, you look for a score where 98% of the population scores below this value and only 2% scores above it. This involves calculating the z-score that corresponds to the 98th percentile and converting it to an actual IQ score.
Using statistical tables or software, you find that the z-score for the 98th percentile is approximately 2.05. The formula to convert a z-score to an IQ score is:
\[ \text{IQ score} = \mu + (z \times \sigma) \]
Plugging in the values:
\[ \text{IQ score} = 100 + (2.05 \times 15) = 130.75 \]
Therefore, someone would need an IQ score of approximately 131 to be eligible for Mensa membership.
Understanding this calculation provides insight into how percentiles and normal distribution are used in real-world applications like IQ testing and high IQ societies.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d390e3f-8275-41b8-b687-67a83e0a6b78%2F4265eb2f-8a90-49a7-be16-3d4086428a0f%2Fnebfuxu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Mensa Membership Requirements**
**Question:**
Mensa, the high IQ society, only accepts membership into their organization if an applicant's IQ is at or above the 98th percentile. If IQ scores follow a normal distribution with a mean of 100 and a standard deviation of 15, what score would someone need in order to be eligible for Mensa membership?
**Explanation:**
To determine eligibility for Mensa based on IQ scores, understanding percentiles and normal distribution is key. The problem specifies that IQ scores are distributed normally, which means they form a bell-shaped curve centered around the mean. Here's a breakdown of the terms:
1. **Mean (μ):** The average score. For IQ, this is 100.
2. **Standard Deviation (σ):** A measure of how spread out the scores are around the mean. For IQ, the standard deviation is 15.
**Finding the Score Required:**
To find the score corresponding to the 98th percentile, you look for a score where 98% of the population scores below this value and only 2% scores above it. This involves calculating the z-score that corresponds to the 98th percentile and converting it to an actual IQ score.
Using statistical tables or software, you find that the z-score for the 98th percentile is approximately 2.05. The formula to convert a z-score to an IQ score is:
\[ \text{IQ score} = \mu + (z \times \sigma) \]
Plugging in the values:
\[ \text{IQ score} = 100 + (2.05 \times 15) = 130.75 \]
Therefore, someone would need an IQ score of approximately 131 to be eligible for Mensa membership.
Understanding this calculation provides insight into how percentiles and normal distribution are used in real-world applications like IQ testing and high IQ societies.
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