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Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
All textbooksTriola 14th EditionCh. 6 - Normal Probability DistributionsProblem 6.R.6a
Chapter 6, Problem 6.R.6a

Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.


a. Find the minimum Wechsler IQ test score that satisfies the Mensa requirement.

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1
Step 1: Understand the problem. Mensa membership requires a score in the top 2% of the population on a standard intelligence test. This means we need to find the IQ score corresponding to the 98th percentile of a normal distribution with a mean (μ) of 100 and a standard deviation (σ) of 15.
Step 2: Recall the properties of the normal distribution. The 98th percentile corresponds to a cumulative probability of 0.98. To find the z-score associated with this percentile, use a z-table or statistical software. The z-score is a standardized value that represents the number of standard deviations a data point is from the mean.
Step 3: Use the z-score formula to find the IQ score. The formula is: x=μ+z×σ, where x is the IQ score, μ is the mean, z is the z-score, and σ is the standard deviation.
Step 4: Substitute the values into the formula. Use the mean (μ = 100), standard deviation (σ = 15), and the z-score obtained from the z-table or software for the 98th percentile.
Step 5: Solve the equation to find the minimum IQ score required for Mensa membership. This will give you the score corresponding to the top 2% of the population.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. In the context of IQ scores, this means that most individuals score around the average (100), with fewer individuals scoring significantly higher or lower. Understanding this concept is crucial for determining percentiles and interpreting scores.
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Z-Score

A Z-score represents the number of standard deviations a data point is from the mean. It is calculated by subtracting the mean from the score and then dividing by the standard deviation. In this case, to find the minimum IQ score for Mensa membership, we need to determine the Z-score that corresponds to the top 2% of the distribution, which helps in calculating the specific IQ score needed.
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Percentiles

Percentiles are measures that indicate the relative standing of a value within a dataset. For example, being in the top 2% means that a score is higher than 98% of the scores in the distribution. To find the minimum IQ score for Mensa, we need to identify the score that corresponds to the 98th percentile of the Wechsler IQ test distribution, which is essential for meeting the membership criteria.
Related Practice
Textbook Question

In Exercises 8 and 9, assume that women have standing eye heights that are normally distributed with a mean of 59.7 in. and a standard deviation of 2.5 in. (based on anthropometric survey data from Gordon, Churchill, et al.).

a. If an eye recognition security system is positioned at a height that is uncomfortable for women with standing eye heights less than 54 in., what percentage of women will find that height uncomfortable?

Textbook Question

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

e. If the mean bone density test score is found for 9 randomly selected subjects, find the probability that the mean is greater than 0.23.

Textbook Question

Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.


b. If 4 randomly selected adults take the Wechsler IQ test, find the probability that their mean score is at least 131.

Textbook Question

Blue Eyes Assume that 35% of us have blue eyes (based on a study by Dr. P. Soria at Indiana University).


c. Find the probability of randomly selecting three different people and finding that all of them have blue eyes.

Textbook Question

In Exercises 8 and 9, assume that women have standing eye heights that are normally distributed with a mean of 59.7 in. and a standard deviation of 2.5 in. (based on anthropometric survey data from Gordon, Churchill, et al.).


Significance Instead of using 0.05 for identifying significant values, use the criteria that a value x is significantly high if P(x or greater) ≤ 0.01 and a value is significantly low if P(x or less) ≤ 0.01. Find the standing eye heights of women that separate significant values from those that are not significant. Using these criteria, is a woman’s standing eye height of 67 in. significantly high?

Textbook Question

In Exercises 1 and 2, use the following wait times (minutes) at 10:00 AM for the Tower of Terror ride at Disney World (from Data Set 33 “Disney World Wait Times” in Appendix B).


35 35 20 50 95 75 45 50 30 35 30 30


e. Convert the longest wait time to a z score.

f. Based on the result from part (e), is the longest wait time significantly high?