Problem 58
Question
Intelligence quotients on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. Intelligence quotients on the Wechsler intelligence test are normally distributed with a mean of 100 and a standard deviation of 15. Use this information to solve Exercises 57-58. Use \(z\)-scores to determine which person has the higher IQ: an individual who scores 150 on the Stanford-Binet or an individual who scores 148 on the Wechsler.
Step-by-Step Solution
Verified Answer
The individual who scores 148 on the Wechsler test has a higher IQ, given the respective standard deviations of the two tests, as demonstrated by the higher Z-score (3.2 vs 3.125).
1Step 1: Compute the z-score for the Stanford-Binet test
To calculate the Z-score, subtract the mean score of the test from the individual's score, and divide the result by the test's standard deviation.The formula is: \(z = (X - μ) / σ\), where \(X\) is the individual's score, \(μ\) is the mean score, and \(σ\) is the standard deviation.\n For the Stanford-Binet test: \(z = (150 - 100) / 16 = 3.125\).
2Step 2: Compute the z-score for the Wechsler test
Using the same formula as before, let's calculate the z-score for the Wechsler test. For the Wechsler test: \(z = (148 - 100) / 15 = 3.2\).
3Step 3: Compare the z-scores
In the final step, compare the two z-scores obtained earlier. The person with a higher z-score has a higher IQ.
Key Concepts
Understanding the Z-ScoreNormal Distribution BasicsThe Importance of Mean and Standard DeviationAn Overview of IQ TestsExploring Stanford-Binet and Wechsler Tests
Understanding the Z-Score
The z-score is a crucial concept in statistics that helps us understand how far, in terms of standard deviations, a particular data point is from the mean. Think of it as a way to standardize scores across different scales.
To calculate it, you use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where
In our exercise, using z-scores helps determine who has a higher score relative to their respective intelligence test's parameters.
To calculate it, you use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where
- \(X\) is the individual score,
- \(\mu\) is the mean,
- and \(\sigma\) is the standard deviation.
In our exercise, using z-scores helps determine who has a higher score relative to their respective intelligence test's parameters.
Normal Distribution Basics
The normal distribution is a common way to represent real-valued random variables. It's often described as a bell curve due to its shape.
This model is significant in statistics because it helps to visually understand data, showing how data points spread around the mean.
Characteristics of a normal distribution include:
This is useful for educational and psychological assessments.
This model is significant in statistics because it helps to visually understand data, showing how data points spread around the mean.
Characteristics of a normal distribution include:
- Symmetry around the mean.
- The mean, median, and mode are all equal.
- About 68% of data falls within one standard deviation of the mean.
This is useful for educational and psychological assessments.
The Importance of Mean and Standard Deviation
Mean and standard deviation are key components in statistical analysis. The mean provides a measure of central tendency, essentially summarizing the dataset with an average value. Standard deviation, on the other hand, shows how much individual data points scatter from the mean. It is calculated as:\[\sigma = \sqrt{\frac{1}{N}\sum{(X_i - \mu)^2}}\] where
They guide statistical inference, prediction, and testing by providing a framework to measure variations and compare results.
- \(N\) is the number of observations,
- \(X_i\) represents each score,
- \(\mu\) is the mean.
They guide statistical inference, prediction, and testing by providing a framework to measure variations and compare results.
An Overview of IQ Tests
IQ tests are standardized assessments designed to measure human intelligence. They give a numerical score which reflects cognitive capabilities relative to others.
Common aspects tested include:
Understanding their distribution, mean, and standard deviation allows for effective interpretation of how individuals perform compared to an average population.
Common aspects tested include:
- logical reasoning,
- problem-solving,
- mathematical skills,
- and verbal knowledge.
Understanding their distribution, mean, and standard deviation allows for effective interpretation of how individuals perform compared to an average population.
Exploring Stanford-Binet and Wechsler Tests
The Stanford-Binet and Wechsler tests are prominent IQ tests used globally. Both are designed to measure intelligence, but they have slight differences in methodology and scoring.
- The **Stanford-Binet** test has a mean of 100 and a standard deviation of 16. It is historically significant, having been one of the first intelligence tests used in modern settings.
- The **Wechsler** test also has a mean of 100 but a smaller standard deviation of 15. It assesses various intelligence areas, providing a comprehensive profile of an individual's cognitive abilities.
These differences in standard deviation mean that while both tests are centered around the same average, the spread of scores differs slightly. This is important when comparing scores between the two, as seen in our exercise: calculating z-scores enables direct comparisons, highlighting individual results proportional to their test distribution.
Other exercises in this chapter
Problem 57
Explain how to use the correlation coefficient for a sample to determine if there is a correlation in the population.
View solution Problem 57
Intelligence quotients on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. Intelligence quotient
View solution Problem 59
A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises \(59-66\), find the data item in this distribution t
View solution Problem 60
A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises \(59-66\), find the data item in this distribution t
View solution