Problem 49 - College Algebra and Calculus: An Applied Approach

Question

A survey of high school seniors from a certain school district who took the SAT has determined that the mean score on the mathematics portion was 650 with a standard deviation of \(12.5\). (a) Assuming the data can be modeled by a normal probability density function, find a model for these data. (b) Use a graphing utility to graph the model. Be sure to choose an appropriate viewing window. (c) Find the derivative of the model. (d) Show that \(f^{\prime}>0\) for \(x<\mu\) and \(f^{\prime}<0\) for \(x>\mu\).

Step-by-Step Solution

Answer

The normal distribution model for these data is \(f(x) = \frac{1}{\sqrt{2\pi(12.5^2)}}e^{-\frac{(x-650)^2}{2(12.5^2)}}\). The derivative is \(f'(x) = f(x) \frac{-(x-\mu)}{\sigma^2}\). It can be shown that \(f^{\prime}>0\) for \(x<\mu\) and \(f^{\prime}<0\) for \(x>\mu\), indicating the function increases before the mean and decreases afterwards.

Step 1: Formulation of the Model

Given that SAT Math scores follow a normal distribution with mean \(\mu = 650\) and standard deviation \(\sigma = 12.5\). The normal distribution model can be written as \[ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{ -\frac{(x-\mu)^2}{2\sigma^2}} \], So substituting the values, we get \[f(x) = \frac{1}{\sqrt{2\pi(12.5^2)}}e^{-\frac{(x-650)^2}{2(12.5^2)}}\]

Step 2: Graphing the Model

In order to graph the function use a graphing utility such as Desmos or others. The viewing window should be adjusted to properly view the distribution. A reasonable area could be [600,700] for x-axis (as it's close but beyond one standard deviation both ways from mean) and [0,0.03] for y-axis since we know that normal distribution tapers off to \(0\).

Step 3: Derive the Model

The derivative of the normal distribution function with respect to \(x\) can be obtained by applying the chain rule and differentiating it piecewise. Hence, it yields to \[f'(x) = f(x) \frac{-(x-\mu)}{\sigma^2}\]

Step 4: Solve for \(f^{\prime}>0\) and \(f^{\prime}<0\)

To show that \(f^{\prime}>0\) for \(x<\mu\) and \(f^{\prime}<0\) for \(x>\mu\), we can plug in a value for \(x\) less than \(\mu (650)\) and greater than \(\mu\) respectively into \(f^{\prime}(x)\). For example, testing with \(x=600\) and \(x=700\) respectively, would yield positive and negative results for \(f^{\prime}(x)\) respectively, thereby validating the conditions. This demonstrates that the function is increasing before the mean and decreasing after that.

Key Concepts

Mean and Standard Deviation

In statistics, the mean and standard deviation are two fundamental concepts used to describe a dataset. The mean is the average of all the data points, which gives us a central value where the data points tend to cluster. In the context of the SAT scores discussed in the exercise, the mean score is 650. This means that most students scored around 650 on the math section.

Standard deviation, on the other hand, quantifies the variation or spread of the data points from the mean. A smaller standard deviation means the data points are closer to the mean, while a larger standard deviation indicates more spread out data. In our exercise, the standard deviation is 12.5, suggesting that most students scored within 12.5 points above or below the mean of 650.

• The mean provides a measure of central tendency.
• Standard deviation measures the spread of the data.

Together, these values are crucial in formulating and understanding the normal distribution model for SAT scores.

Probability Density Function

A probability density function (PDF) is a function that describes the likelihood of obtaining the possible values that a random variable can take. For our SAT score example, the normal distribution is a type of PDF that is symmetrical and has a distinctive bell curve shape.

The PDF for a normal distribution is defined as: \[f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\]Here,

• \(\mu\) is the mean.
• \(\sigma\) is the standard deviation.
• \(x\) represents the variable (in this case, SAT scores).

The curve has its peak at the mean, and it is symmetric around this central point. The tails of the curve extend infinitely without touching the x-axis, attributing probabilities to all potential scores.
Understanding the PDF helps in visualizing the distribution of scores and assessing how likely it is for a score to occur within a certain range.

Graphing Mathematical Functions

Graphing mathematical functions like the normal distribution allows us to visually analyze data and understand behavior. When graphing our PDF model for the SAT scores, ensuring an appropriate viewing window is crucial. This involves selecting intervals for the x-axis and y-axis that showcase the function clearly.

For the SAT score distribution:

• An x-axis range of [600,700] covers the scores within one standard deviation from the mean.
• Setting the y-axis from [0,0.03] accommodates the peak probability density.

Using tools such as Desmos or graphing calculators simplifies this process. Accurate graphing helps illustrate how scores are distributed; it's clear that most scores cluster around the mean, decreasing as they move away.
Remember, graphing offers an intuitive look at complex mathematical relationships and confirms theoretical calculations.

Chain Rule in Calculus

The chain rule is an essential calculus tool for differentiating composite functions. In the context of our normal distribution model, we use the chain rule to find the derivative of the probability density function (PDF). This derivative helps us understand the behavior of the distribution—how it changes or varies at different points.

The derivative of the normal distribution \( f(x) \) with respect to \( x \) is given by the formula:\[f'(x) = f(x)\frac{-(x-\mu)}{\sigma^2}\]In applying the chain rule:

• Differentiate the exponential component \(-\frac{(x-\mu)^2}{2\sigma^2}\).
• Multiply it by the existing function \(f(x)\).

For the SAT scores, evaluating conditions for \(f^{\prime}(x)>0\) when \(x<\mu\) and \(f^{\prime}(x)<0\) when \(x>\mu\) illustrates that the curve rises until the mean and falls afterward, demonstrating increase and decrease phases in the score distribution around the mean.