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$$ \begin{array}{c}{\text { The family of bell-shaped curves }} \\\ {y=\frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-\mu)^{2} / / 2 \sigma^{2})}}\end{array} $$ $$ \begin{array}{l}{\text { occurs in probability and statistics, where it is called the }} \\ {\text { normal density function. The constant } \mu \text { is called the mean }} \\ {\text { and the positive constant } \sigma \text { is called the standard deviation. }}\end{array} $$ $$ \begin{array}{c}{\text { For simplicity, let's scale the function so as to remove the }} \\ {\text { factor } 1 /(\sigma \sqrt{2 \pi}) \text { and let's analyze the special case where }} \\ {\mu=0 . \text { So we study the function }} \\ {f(x)=e^{-x^{2} /\left(2 \sigma^{2}\right)}}\end{array} $$ $$ \begin{array}{l}{\text { (a) Find the asymptote, maximum value, and inflection }} \\ {\text { points of } f .} \\ {\text { (b) What role does } \sigma \text { play in the shape of the curve? }} \\ {\text { (c) Illustrate by graphing four members of this family on }} \\ {\text { the same screen. }}\end{array} $$

Short Answer

Expert verified
The maximum of \( f(x) \) is 1 at \( x = 0 \), inflection points are at \( x = \pm \sigma \), and \( y = 0 \) is an asymptote. \( \sigma \) affects the width of the curve.

Step by step solution

01

Understanding the Asymptote

The function \( f(x) = e^{-x^2 / (2 \sigma^2)} \) is an exponential function. We look for horizontal asymptotes by considering \( \lim_{x \to \pm\infty} f(x) = 0 \). Therefore, \( y = 0 \) is a horizontal asymptote.
02

Finding the Maximum Value and Its Location

The function \( f(x) \) is symmetric around \( x = 0 \). We find the derivative \( f'(x) = -\frac{x}{\sigma^2} e^{-x^2/(2\sigma^2)} \). Setting \( f'(x) = 0 \) leads to \( x = 0 \) being a critical point. Since \( f''(0) < 0 \), 0 is a maximum and \( f(0) = 1 \) is the maximum value.
03

Determining Inflection Points

The second derivative is \( f''(x) = \left(\frac{x^2 - \sigma^2}{\sigma^4}\right) e^{-x^2/(2\sigma^2)} \). Setting \( f''(x) = 0 \), we solve \( x^2 = \sigma^2 \), resulting in inflection points at \( x = \pm \sigma \).
04

Exploring the Role of \( \sigma \)

The parameter \( \sigma \) influences the spread of the curve. Larger \( \sigma \) values result in wider curves, while smaller \( \sigma \) values lead to narrower curves. It effectively determines the 'width' of the bell shape.
05

Graphing the Family of Curves

To illustrate the effect of \( \sigma \), plot \( f(x) = e^{-x^2/(2\sigma^2)} \) for different \( \sigma \) values, such as \( \sigma = 0.5, 1, 1.5, 2 \). Observing the plots, we see that as \( \sigma \) increases, the curve becomes flatter and wider.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
The standard deviation, represented by the symbol \( \sigma \), is a measure of how spread out the values in a data set are. In the context of the normal distribution, it helps determine the shape of the curve.

Think of the standard deviation as a measure of variability. A small \( \sigma \) means the data points are close to the mean, resulting in a steeper and narrower curve, or bell shape. Conversely, a larger \( \sigma \) indicates that data points are more spread out, thus the curve becomes wider and flatter.

**Role of \( \sigma \) in Shape:**
  • Larger \( \sigma \): Wider and flatter curve
  • Smaller \( \sigma \): Steeper and narrower curve
Understanding this principle is essential when analyzing data distributions as it provides insight into the 'spread' of data around a mean value.
Mean
Within the framework of a normal distribution, the mean, denoted by \( \mu \), acts as the center of the distribution. It's the point around which data values are symmetrically distributed. The simplicity of the symmetric property makes the mean a crucial concept in statistics.

The mean can be thought of as the balance point of the distribution. In our simplified function \( f(x) = e^{-x^2 /(2 \sigma^2)} \), we set \( \mu = 0 \). Hence, the peak of the curve is centered at zero, and the distribution spreads out from this central point.

**Key Characteristics of the Mean:**
  • It determines the central tendency of the distribution
  • In a normal distribution, the mean, median, and mode coincide
  • For symmetrical data, the mean lies at the center
Understanding how the mean functions helps in interpreting where the average or most common values in a dataset would be.
Exponential Function
An exponential function is a mathematical function that features an exponent and undergoes rapid growth or decay. In our function \( f(x) = e^{-x^2/(2\sigma^2)} \), the exponential term \( e^{-x^2/(2\sigma^2)} \) dictates the bell-shaped curve characteristic of the normal distribution.

This specific use of an exponential function results in a rapid decrease of values as \( x \) moves away from the mean. As a result, it forms the classic 'bell shape' synonymous with normal distributions. The base of the exponent, \( e \), is the natural base approximately equal to 2.718, which is critical for defining this smooth, continuous curve.

**Characteristics of Exponential Functions in Normal Distribution:**
  • Results in rapid decay or growth based on the exponent's sign
  • Smoothly transitions values, creating the continuous curve
  • Center-focused (at \( \mu = 0 \) for our function), symmetrically spreading outwards
By understanding the exponential function, students can decipher how this rapid decrease generates the typical curve seen in statistics.
Inflection Points
Inflection points occur at values of \( x \) where the concavity of a function changes. For the function \( f(x) = e^{-x^2 /(2 \sigma^2)} \), the inflection points are identified by solving \( f''(x) = 0 \). These points are critical as they indicate where the curve shifts from being concave up (curving upwards) to concave down (curving downwards), or vice versa.

In our function, resulting inflection points are at \( x = \pm \sigma \). This implies:
  • To the left of \( -\sigma \) and to the right of \( \sigma \), the curve is concave down
  • Between \( -\sigma \) and \( \sigma \), the curve is concave up
This switch in curvature is what gives the smoothly transitioning slope on either side of the peak of a normal distribution.

**Understanding Inflection Points:**
The concept of inflection points helps dissect the curvature characteristics of the distribution and explains the seamless shift in the direction of the curve's slope.

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Most popular questions from this chapter

Age and size at maturity Most organisms grow for a period of time before maturing reproductively. For many species of insects and fish, the later the age \(a\) at maturity, the larger the individual will be, and this translates into a greater reproductive output. At the same time, however, the probability of surviving to maturity decreases as the age of maturity increases. These contrasting effects can be combined into a single measure of reproductive success in different ways. Suppose \(\mu\) is a constant representing mortality rate. Find the optimal age at maturity for the following models. (a) \(r=\frac{\ln \left(a e^{-\mu a}\right)}{a}\) (b) \(R=a e^{-\mu a}\)

$$ \begin{array}{l}{\text { Suppose } f(3)=2, f^{\prime}(3)=\frac{1}{2}, \text { and } f^{\prime}(x)>0 \text { and } f^{\prime \prime}(x)<0} \\ {\text { for all } x .} \\ {\text { (a) Sketch a possible graph for } f \text { . }} \\\ {\text { (b) How many solutions does the equation } f(x)=0 \text { have? }} \\\ {\text { Why? }} \\ {\text { (c) Is it possible that } f^{\prime}(2)=\frac{1}{3} ? \text { Why? }}\end{array} $$

\(37-44\) (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of \(f .\) $$f(x)=\ln (1-\ln x)$$

$$ \begin{array}{l}{\text { Mutation accumulation } \text { When a population is subjected }} \\ {\text { to a mutagen, the fraction of the population that contains }} \\ {\text { at least one mutation increases with the duration of the }} \\ {\text { exposure. A commonly used equation describing this }} \\\ {\text { fraction is } f(t)=1-e^{-\mu t}, \text { where } \mu \text { is the mutation rate }} \\ {\text { and is positive. Suppose we have two populations, A and B. }}\end{array} $$. $$ \begin{array}{l}{\text { Population A is subjected to the mutagen for } 3 \text { min }} \\ {\text { whereas, with population } \mathrm{B} \text { , half of the individuals are }} \\ {\text { subjected to the mutagen for } 2 \text { min and the other half for }} \\ {4 \text { min. Which population will have the largest fraction of }} \\ {\text { mutants? Explain your answer using derivatives. }}\end{array} $$

Solve the initial-value problem. \(\frac{d u}{d t}=\sqrt{t}+\frac{2}{\sqrt{t}}, \quad t>0, u(1)=5\)

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