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?

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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.
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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