Problem 50 - College Algebra and Calculus: An Applied Approach

Question

A survey of a college freshman class has determined that the mean height of females in the class is 64 inches with a standard deviation of \(3.2\) inches. (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 probability density function for these data is \(f(x) = \frac{1}{3.2\sqrt{2\pi}} e^{ -\frac{1}{2} ((x - 64) / 3.2)^2 }\). The derivative is \(f^{\prime}(x) = \frac{-1}{10.24\sqrt{2\pi}} (x-64) e^{ -\frac{1}{2} ((x - 64) / 3.2) ^2}\), and \(f^{\prime}>0\) for \(x<64\) and \(f^{\prime}<0\) for \(x>64\)

Step 1: Creating the Probability Density Function

The general form of a normal probability density function is \(f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{ -\frac{1}{2} ((x - \mu) / \sigma)^2 }\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation. In this case, \(\mu = 64\), and \(\sigma = 3.2\), so our function becomes: \(f(x) = \frac{1}{3.2\sqrt{2\pi}} e^{ -\frac{1}{2} ((x - 64) / 3.2)^2 }\)

Step 2: Graph the Model

A graphing utility can be used to plot this function. We can't show graphics in this environment. However, an appropriate viewing window would likely include heights from 60 to 68 inches on the x-axis and probabilities from 0 to 0.15 on the y-axis.

Step 3: Find the Derivative of the Model

The derivative of our function with respect to x is: \(f^{\prime}(x) = \frac{d}{dx} \left[\frac{1}{3.2\sqrt{2\pi}} e^{ -\frac{1}{2} ((x - 64)/3.2) ^2} \right] = \frac{-1}{10.24\sqrt{2\pi}} (x-64) e^{ -\frac{1}{2} ((x - 64) / 3.2) ^2}\)

Step 4: Analyze the Derivative

The derivative \(f^{\prime}(x)\) is positive for \(x < 64\) and negative for \(x > 64\) as the factor \((x-64)\) is negative for \(x < 64\) and positive for \(x > 64\), while the exponential part is always positive for real \(x\). This proves that \(f^{\prime}>0\) for \(x<64\) and \(f^{\prime}<0\) for \(x>64\)

Key Concepts

Probability Density Function

A Probability Density Function, often abbreviated as PDF, is a crucial concept in statistics, especially when discussing normal distribution. It serves as a mathematical function that indicates the probability of a random variable to take on a given value. Probability Density Functions are used to model and work with continuous probability distributions.

In this exercise, we make use of the PDF of a normal distribution, which captures how the data are distributed around a mean. The formula for a normal probability density function is given by:
\[f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{ -\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2 }\]
Here, \(\mu\) represents the mean of the dataset, \(\sigma\) is the standard deviation, and \(e\) is the base of the natural logarithm.

In our case, the mean height \(\mu\) is 64 inches and the standard deviation \(\sigma\) is 3.2 inches. By plugging these values into our PDF formula, we obtain a model:
\[f(x) = \frac{1}{3.2\sqrt{2\pi}} e^{ -\frac{1}{2} \left(\frac{x - 64}{3.2}\right)^2 }\]

This function provides us with a bell-shaped curve that reflects how most female students' heights cluster around 64 inches. The total area under the curve is 1, which represents the total probability.

Standard Deviation

Standard deviation is a statistic that explains how dispersed the values in a dataset are relative to the mean. It's critical in understanding the spread of the dataset. The formula for the standard deviation \(\sigma\) in a sample is:
\[\sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^{N} (x_i - \mu)^2}\]
where \(x_i\) are the data points, \(\mu\) is the mean, and \(N\) is the sample size.

In the context of our exercise, the standard deviation is 3.2 inches. This indicates the typical amount by which female heights deviate from the mean height of 64 inches. A smaller standard deviation would suggest that the heights are more consistent or clustered closely around the mean, whereas a larger standard deviation would mean more variability.

Understanding standard deviation is pivotal when interpreting the normal distribution since it influences the width of the bell curve. Approximately 68% of data within a normally distributed dataset lies within one standard deviation from the mean, reinforcing how the standard deviation frames the expectation around the mean in a dataset.

Derivative of a Function

Derivatives are a fundamental concept in calculus, letting us understand how a function changes as its input changes. In simple terms, the derivative gives us the slope of the function at any given point. This is especially useful in understanding rates of change and how a process is evolving over time or other parameters.

For our given probability density function, the derivative \(f^{\prime}(x)\) tells us how the probability density changes as the height \(x\) changes. The derivative of a normally distributed function with respect to \(x\) is:
\[f^{\prime}(x) = \frac{-1}{10.24\sqrt{2\pi}} (x-64) e^{ -\frac{1}{2} \left(\frac{x - 64}{3.2}\right)^2}\]
From the derivative, we can infer that:

• When \(x < 64\), \(f^{\prime}(x)\) is positive, suggesting the slope of the PDF is ascending as it moves towards the mean.
• When \(x > 64\), \(f^{\prime}(x)\) is negative, indicating the descending slope as it moves away from the mean.

This means precisely at the mean (\(x=64\)), the function neither increases nor decreases, being at the peak of the bell curve. Understanding derivatives in this context helps us recognize where the function increases and decreases, a powerful tool in mapping out behavior of real-world events.