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Chapter 3: Problem 7

Show that the graph of a pdf \(N\left(\mu, \sigma^{2}\right)\) has points of inflection at \(x=\mu-\sigma\) and \(x=\mu+\sigma\).

Short Answer

Expert verified
The graph of a standard normal distribution pdf \(N(\mu, \sigma^{2})\) has points of inflection at \(x=\mu-\sigma\) and \(x=\mu+\sigma\). This is proven by calculating the second derivative of the function and setting it equal to zero. The solutions of this equation correspond to the x-coordinates of the points of inflection on the graph of the function.

Step by step solution

01

Write down the function

First, write down the probability density function for a Normal Distribution in the following manner: \(f(x) = \frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}\).
02

Find the second derivative

Find the second derivative \(f''(x)\) of the function. The second derivative measures the rate at which the derivative \(f'(x)\) changes. The points of inflection of the function can be found by setting the second derivative equal to zero.
03

Solve the second derivative equation for x

After calculating the second derivative, set it equal to zero and find the roots, or values of x, by solving the equation. These roots represent the x-coordinates of the points of inflection.
04

Verify the points of inflection

Substitute the values of x, \(x=\mu-\sigma\) and \(x=\mu+\sigma\), into the second derivative function to check for the change in the sign of the second derivative. If the second derivative changes its sign at these points, then these points would indeed be the points of inflection.

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Most popular questions from this chapter

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