Problem 10 - 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

(a) The density function is \( f(x) = \frac{1}{15\sqrt{2\pi}} e^{-\frac{1}{2} \left( \frac{x-100}{15} \right)^2 } \). (b) About 6.69% of the population has an IQ between 115 and 120.

Step 1: Understanding the Normal Distribution Formula

The normal distribution is described by its probability density function (PDF). Given that the mean \( \mu = 100 \) and the standard deviation \( \sigma = 15 \), the PDF is given by:\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2}\]Substituting the given values:\[f(x) = \frac{1}{15 \sqrt{2\pi}} e^{-\frac{1}{2} \left( \frac{x-100}{15} \right)^2}\]

Step 2: Computing Z-Scores

To find the probability of IQ scores between 115 and 120, we first compute the Z-scores. The Z-score formula is:\[ Z = \frac{x - \mu}{\sigma} \]For \( x = 115 \),\[ Z = \frac{115 - 100}{15} = 1.00 \]For \( x = 120 \),\[ Z = \frac{120 - 100}{15} = 1.33 \]

Step 3: Finding Probabilities Using Z-table

Using the Z-table, we find the probability of a Z-score being less than a certain value:\( P(Z < 1.00) \) and \( P(Z < 1.33) \).From the Z-table, \( P(Z < 1.00) \approx 0.8413 \) and \( P(Z < 1.33) \approx 0.9082 \).

Step 4: Calculating the Fraction of the Population

To find the fraction of the population with IQ between 115 and 120, subtract the two probabilities:\[ P(1.00 < Z < 1.33) = P(Z < 1.33) - P(Z < 1.00) = 0.9082 - 0.8413 = 0.0669 \]

Step 5: Interpreting the Result

The resulting value indicates that approximately 6.69% of the population has an IQ between 115 and 120 based on the normal distribution model.

Key Concepts

Probability Density Function

In the heart of understanding the normal distribution is the concept of the Probability Density Function (PDF). In simple terms, the PDF helps us describe how the probabilities are distributed over the values of a random variable. For a normal distribution, this is represented by a specific bell-shaped curve.

The formula for the normal distribution PDF is:

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

Here, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. The equation ensures that all probabilities add up to 1, meaning it accounts for all possible outcomes. By substituting the mean (\( \mu = 100 \)) and standard deviation (\( \sigma = 15 \)), you can determine the probability of any IQ score using the corresponding PDF formula.

The Gaussian curve shows that most IQ scores will cluster around the mean, with fewer at the extremes. Understanding this layout helps in predicting how likely an individual score is.

Z-scores

Z-scores are a standardized measure that express a value's distance from the mean in units of standard deviation. They are central to comparing individual values against a common reference in a normal distribution.

To calculate a Z-score, use the formula:

• \( Z = \frac{x - \mu}{\sigma} \)

This tells us how many standard deviations an element \( x \) is from the mean \( \mu \).

For example, if we are calculating the Z-scores for IQ scores of 115 and 120 where the mean IQ is 100 and standard deviation is 15, we find:

• For 115: \( Z = \frac{115 - 100}{15} = 1.00 \)
• For 120: \( Z = \frac{120 - 100}{15} = 1.33 \)

These Z-scores indicate the position of the specific IQ scores relative to the mean IQ score.

Z-table

The Z-table is an essential tool for finding probabilities associated with Z-scores in a standard normal distribution. It allows you to determine the probability that a standard normal variable will be less than or equal to a given Z-score.

When you compute a Z-score, you can use the Z-table to determine what proportion of the data falls below that score.
For instance, with Z-scores of 1.00 and 1.33:

• The Z-table shows that \( P(Z < 1.00) \approx 0.8413 \)
• For \( P(Z < 1.33) \approx 0.9082 \)

Subtracting these gives the probability of values being between these Z-scores.

• \( P(1.00 < Z < 1.33) = 0.9082 - 0.8413 = 0.0669 \)

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

Standard Deviation

Standard Deviation is a measure of the amount of variation or dispersion in a set of values. In a normal distribution, it illustrates how spread out the numbers are around the mean.

For example, in our IQ distribution, a standard deviation of 15 means that most of the IQ scores will fall within 15 points of the average IQ of 100. The larger the standard deviation, the broader the distribution and the greater the spread of the values.

• Standard deviation not only helps in determining where most values lie but also in calculating the Z-scores, by providing the scaling factor for the difference between the value and mean.
• By recognizing the importance of standard deviation, one can better understand why certain scores deviate significantly from the average.

In conclusion, understanding standard deviation is crucial to interpreting how clustered or dispersed data is in the context of normal distribution.