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Chapter 5: Problem 10

People with \(z\) -scores above 2.5 on an IQ test are sometimes classified as geniuses. If IQ scores have a mean of 100 and a standard deviation of 16 points, what IQ score do you need to be considered a genius?

Short Answer

Expert verified
An IQ score of 140 is needed to be considered a genius.

Step by step solution

01

Understanding the Problem

We know that someone is classified as a genius if their z-score on an IQ test is above 2.5. Given that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16 points, we are tasked with finding the IQ score corresponding to a z-score of 2.5.
02

Z-score Formula

The z-score indicates how many standard deviations an element is from the mean. The formula to calculate the z-score is: \( z = \frac{x - \mu}{\sigma} \), where \(z\) is the z-score, \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
03

Setting Up the Equation

We are given a z-score of 2.5, a mean \(\mu = 100\), and a standard deviation \(\sigma = 16\). We only need to compute the IQ score \(x\) using the z-score formula: \( 2.5 = \frac{x - 100}{16} \).
04

Solving for the IQ Score

To find \(x\), we can rearrange the equation: \[ x - 100 = 2.5 \times 16 \]. Then solve for \(x\): \[ x = 2.5 \times 16 + 100 \].
05

Computing the Final Answer

Calculate the multiplication: \[ 2.5 \times 16 = 40 \]. Add this to the mean: \[ x = 40 + 100 = 140 \].
06

Conclusion

An IQ score of 140 or greater is needed to be classified as a genius according to the given criteria.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-Score
In statistics, a Z-score tells us how many standard deviations a data point is from the mean of the dataset. It helps compare individual scores to the overall distribution, usually in a normal distribution. Here鈥檚 how it works:
  • The Z-score is calculated using the formula: \( z = \frac{x - \mu}{\sigma} \), where:
  • \( z \) is the Z-score.
  • \( x \) is the score being examined.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation of the dataset.
A high Z-score means the score is far above the mean, while a low Z-score indicates it is below it. In the context of IQ tests, having a Z-score above 2.5 signifies a much higher IQ compared to the average person.
Standard Deviation
Standard deviation is a key concept to understanding the spread or dispersion of a dataset. It tells us how much the individual values in a dataset differ from the mean.
  • The smaller the standard deviation, the closer the data points are to the mean.
  • A larger standard deviation indicates that the data points are spread out over a wider range of values.
In the case of IQ tests, a standard deviation of 16 points means that the test results vary on average by 16 IQ points from the mean. Understanding standard deviation helps us interpret Z-scores more effectively.
IQ Test
IQ, or Intelligence Quotient, tests are designed to measure human intelligence. They quantify cognitive abilities relative to an average score, usually set at 100.
  • IQ scores follow a normal distribution, with most people scoring near the average.
  • The spread or distribution of scores is expressed in terms of standard deviation.
These tests are valuable for assessing cognitive development and potential. In academic and psychological settings, understanding how IQ scores are distributed can provide insight into an individual's intellectual standing.
Mean
The mean, often known as the average, is a central concept in statistics. It is calculated by adding all the numbers in a dataset and then dividing by the total amount of numbers.
  • In normal distributions, the mean often represents the center point of the data.
  • For IQ scores, the mean is typically set at 100, reflecting the central tendency of intelligence in the population.
The mean is vital when calculating Z-scores and understanding where an individual score falls in the overall data distribution. It provides a baseline for comparison with other data points.

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Most popular questions from this chapter

A company that manufactures rivets believes the shear strength (in pounds) is modeled by \(N(800,50)\) a) Draw and label the Normal model. b) Would it be safe to use these rivets in a situation requiring a shear strength of 750 pounds? Explain. c) About what percent of these rivets would you expect to fall below 900 pounds? d) Rivets are used in a variety of applications with varying shear strength requirements. What is the maximum shear strength for which you would feel comfortable approving this company's rivets? Explain your reasoning.

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of \(188 \mathrm{mg} / \mathrm{dL}\) and a standard deviation of 24 a) Draw and label the Normal model. b) What percent of adult women do you expect to have cholesterol levels over \(200 \mathrm{mg} / \mathrm{dL} ?\) c) What percent of adult women do you expect to have cholesterol levels between 150 and \(170 \mathrm{mg} / \mathrm{dL} ?\) d) Estimate the IQR of the cholesterol levels. e) Above what value are the highest \(15 \%\) of women's cholesterol levels?

The Virginia Cooperative Extension reports that the mean weight of yearling Angus steers is 1152 pounds. Suppose that weights of all such animals can be described by a Normal model with a standard deviation of 84 pounds. a) How many standard deviations from the mean would a steer weighing 1000 pounds be? b) Which would be more unusual, a steer weighing 1000 pounds or one weighing 1250 pounds?

Anna, a language major, took final exams in both French and Spanish and scored 83 on each. Her roommate Megan, also taking both courses, scored 77 on the French exam and 95 on the Spanish exam. Overall, student scores on the French exam had a mean of 81 and a standard deviation of \(5,\) and the Spanish scores had a mean of 74 and a standard deviation of \(15 .\) a) To qualify for language honors, a major must maintain at least an 85 average for all language courses taken. So far, which student qualifies? b) Which student's overall performance was better?

The winning scores of all college men's basketball games for the \(2011-12\) season were approximately normally distributed with mean 77.5 points and standard deviation 12.5 points. a) Draw the Normal model for winning scores. b) What interval of winning scores would be the central \(95 \%\) of all winning scores for the \(2011-12\) season? c) About what percent of the winning scores should be less than 65 points? d) About what percent of the winning scores should be between 65 and 102 points? e) About what percent of the winning scores should be over 102 points?
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