Problem 4 - Applied Calculus

Question

The distribution of IQ scores can be modeled by a normal distribution with mean 100 and standard deviation \(15 .\) (a) Write the formula for the density function of IQ scores. (b) Estimate the fraction of the population with IQ between 115 and 120 .

Step-by-Step Solution

Answer

Approximately 6.69% of the population has an IQ between 115 and 120.

Step 1: Understand the Problem

We are given that IQ scores follow a normal distribution with a mean of 100 and a standard deviation of 15. We need to write the formula for the probability density function (PDF) for this distribution and calculate the fraction of the population with IQ between 115 and 120.

Step 2: Write the Formula for the PDF

For a normally distributed random variable with mean \(\mu\) and standard deviation \(\sigma\), the probability density function is given by: \[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} \] Substitute \(\mu = 100\) and \(\sigma = 15\) into the formula, we have: \[ f(x) = \frac{1}{15 \sqrt{2\pi}} e^{-\frac{(x - 100)^2}{2 \times 15^2}} \] This is the PDF for the IQ scores.

Step 3: Standardize the Range

In order to calculate the fraction of the population with IQ between 115 and 120, we need to standardize these scores using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] For \(X = 115\), the Z-score is \(Z = \frac{115 - 100}{15} = 1\). For \(X = 120\), the Z-score is \(Z = \frac{120 - 100}{15} \approx 1.33\).

Step 4: Use the Standard Normal Distribution Table

Look up the Z-scores in the standard normal distribution table to find probabilities: - \(P(Z \leq 1)\) corresponds to about 0.8413. - \(P(Z \leq 1.33)\) corresponds to about 0.9082. These probabilities give us the areas under the curve to the left of the Z-scores.

Step 5: Calculate the Fraction of Population

To find the fraction of the population with IQs between 115 and 120, subtract the probabilities: \[ P(115 < X < 120) = P(Z < 1.33) - P(Z < 1) = 0.9082 - 0.8413 = 0.0669 \] Thus, approximately 6.69% of the population has an IQ between 115 and 120.

Key Concepts

Probability Density Function

In statistics, the probability density function (PDF) of a continuous random variable describes the likelihood of the variable taking on a certain value. For the normal distribution, the PDF is of particular significance. It helps understand how values are spread around the mean, providing the bell-shaped curve characteristic of the normal distribution.

For any normally distributed random variable with mean \( \mu \) and standard deviation \( \sigma \), the probability density function is expressed mathematically as:

• \( f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} \)

This equation comprises components that reflect the mean and variability of the data. The section \( -\frac{(x - \mu)^2}{2 \sigma^2} \) in the exponent ensures that as the value \( x \) moves away from the mean \( \mu \), the probability decreases, creating the familiar normal curve.The formula is adjusted to specific distributions by substituting the values for \( \mu \) and \( \sigma \). For instance, in the IQ score example, it is adjusted to a mean of 100 and a standard deviation of 15:

• \( f(x) = \frac{1}{15 \sqrt{2\pi}} e^{-\frac{(x - 100)^2}{2 \times 15^2}} \)

Understanding the PDF provides insight into the data distribution and aids in performing further statistical calculations, such as finding probabilities of specific ranges.

Z-score

A Z-score is a statistical measurement that describes a value's position in relation to the mean of a group of values. It expresses how far away a particular data point is from the mean in terms of standard deviations.

To determine a Z-score, you use the formula:

• \( Z = \frac{X - \mu}{\sigma} \)

Where: - \( X \) is the data point,- \( \mu \) is the mean of the data,- \( \sigma \) is the standard deviation.
This standardized score answers the question: "How many standard deviations is \( X \) from the mean?"For example, using the IQ scores problem:

• To find the Z-score for an IQ of 115: \( Z = \frac{115 - 100}{15} = 1 \)
• For an IQ of 120: \( Z = \frac{120 - 100}{15} \approx 1.33 \)

Z-scores are crucial as they allow comparison between different datasets and make it possible to use standard normal distribution tables to find probabilities associated with specific ranges.

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution that has a mean of 0 and a standard deviation of 1. It is a crucial concept because it standardizes values across different datasets, allowing for direct comparison.

The transformation from a normal distribution to a standard normal distribution involves translating values into Z-scores:

• These Z-scores are then plotted on the standard normal distribution curve.

The probability of a value or interval under this distribution is represented by the area under the curve.

In practice, you use a Z-table, or standard normal distribution table, to look up Z-scores and determine the probability that a given Z-score lies below it. For example, a Z-score of 1 corresponds to approximately 0.8413; thus, 84.13% of the data lies below this point.

In the IQ example, the probability that someone's IQ falls between 115 and 120 can be calculated by finding the areas corresponding to Z-scores of 1 and 1.33 and subtracting them, yielding the solution 0.0669 or 6.69%. The ability to use the standard normal distribution in this way makes it a powerful tool in statistics.