Question

The intelligence quotient (IQ) score, as measured by the Stanford-Binet IQ test, is normally distributed in a certain population of children. The mean IQ score is 100 points, and the standard deviation is 16 points. \(^{18}\) What percentage of children in the population have IQ scores (a) 116 or more? (b) 84 or less? (c) between 84 and \(116 ?\) (d) between 84 and \(148 ?\) (e) between 116 and \(148 ?\)

Short answer

Percentage of children with (a) IQ 116 or more is approximately 15.87%, (b) IQ 84 or less is approximately 15.87%, (c) IQ between 84 and 116 is approximately 68.26%, (d) IQ between 84 and 148 is approximately 99.83%, (e) IQ between 116 and 148 is approximately 31.57%.

Step-by-step solution

Understand the Z-score

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It indicates how many standard deviations an element is from the mean. Calculate the Z-score using the formula: Z = (X - mean) / SD, where X is the value, mean is the average, and SD is the standard deviation.

Calculate Z-scores for given IQ scores

To find the percentage of children scoring at various ranges, calculate the Z-scores for the boundaries of the ranges using the provided mean and SD. For example, for 116 or more, Z = (116 - 100) / 16 = 1.

Use Z-table or calculator for probabilities

Consult a Z-table or use a statistical calculator to find the probability corresponding to each Z-score. This probability represents the percentage of scores that fall below the given value.

Find the percentage for IQ 116 or more

For Z = 1, the Z-table gives a probability of 0.8413. To find the percentage scoring above it, subtract this value from 1, and multiply by 100. This gives the percentage of children who have an IQ of 116 or more.

Find the percentage for IQ 84 or less

Calculate Z for IQ 84: Z = (84 - 100) / 16 = -1. For Z = -1, we find a probability of 0.1587 from the Z-table. Multiply this value by 100 to get the percentage of children who have an IQ of 84 or less.

Find the percentage for IQ between 84 and 116

Subtract the probability of scoring less than 84 from the probability of scoring less than 116. This represents the percentage of children with an IQ score between 84 and 116.

Find the percentage for IQ between 84 and 148

Similarly, calculate Z for 148: Z = (148 - 100) / 16 = 3. The probability for Z = 3 is nearly 0.9987. Subtract the probability of a score less than 84 from 0.9987 to get the percentage for IQ between 84 and 148.

Find the percentage for IQ between 116 and 148

Subtract the probability of scoring less than 116 from the probability of scoring less than 148. This will give the percentage of children whose IQ is between 116 and 148.

Key concepts

Z-score Calculation

Understanding Z-score calculations is essential when dealing with normally distributed data, such as IQ scores. A Z-score quantifies how many standard deviations a data point, such as an individual's IQ score, is from the population mean. To calculate the Z-score, use the formula:
\[ Z = \frac{{X - \text{{mean}}}}{{\text{{SD}}}} \]
where \(X\) represents the value in question, the mean is the average of the data set, and SD stands for standard deviation. In our exercise, calculating the Z-score for an IQ of 116 with a mean of 100 and a standard deviation of 16 would be:
\[ Z = \frac{{116 - 100}}{{16}} = 1 \]
As this falls into a measurable range on the standard normal distribution, we can interpret this Z-score using the Z-table to assess probabilities, a crucial step in statistical analysis.

Stanford-Binet IQ Test Analysis

The Stanford-Binet IQ test is designed to measure human intelligence. The test scores are set to follow a normal distribution, with a mean (average) score established at 100 and a standard deviation that gauges the variation in scores. In our exercise, a standard deviation of 16 points is given, which means that most individuals' IQs will fall within 16 points of the mean score. By calculating the Z-scores for specific IQ values, we can use the properties of the normal distribution to interpret how an individual's IQ compares to the broader population. For instance, a Z-score of 1, which corresponds to an IQ of 116, indicates an IQ greater than approximately 84% of the population, given the cumulative nature of the normal distribution probabilities.

Normal Distribution Probability

The normal distribution, known for its symmetrical bell-shaped curve, is key in statistical analysis. It suggests that data points are most frequently observed near the mean, and the likelihood of observing a value decreases as you move further away from the mean. Probabilities in a normal distribution are found using Z-scores. Once you have the Z-score, you can consult a Z-table or use statistical software to find the cumulative probability that a data point will fall below a certain value. In our exercise, to calculate the probability of children with an IQ between 84 and 116, which are Z-scores of -1 and 1, respectively, you would find the cumulative probability for both Z-scores and subtract them to find the probability for the desired IQ range.