Question

complete the statement below
the points at ( x = )__ and ( x = )__ are the inflection points on the normal curve
what are the two points?
a. the points are ( x=mu - 3sigma ) and ( x=mu + 3sigma ).
b. the points are ( x=mu - sigma ) and ( x=mu + sigma ).
c. the points are ( x=mu - 2sigma ) and ( x=mu + 2sigma ).

Explanation:

The inflection points of a normal curve \(y = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}\) occur where the second - derivative changes sign.

1. First, find the first derivative using the chain rule.

Let \(u=-\frac{(x - \mu)^2}{2\sigma^2}\), then \(y=\frac{1}{\sigma\sqrt{2\pi}}e^{u}\).
\(y^\prime=\frac{1}{\sigma\sqrt{2\pi}}e^{u}\cdot u^\prime\), and \(u^\prime =-\frac{x - \mu}{\sigma^2}\), so \(y^\prime=-\frac{x - \mu}{\sigma^3\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}\).

1. Then, find the second derivative using the product rule \((uv)^\prime = u^\prime v+uv^\prime\).

Let \(u =-(x - \mu)\) and \(v=\frac{1}{\sigma^3\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}\).
\(u^\prime=- 1\), \(v^\prime=-\frac{x - \mu}{\sigma^5\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}\).
\(y^{\prime\prime}=\frac{1}{\sigma^3\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}-\frac{(x - \mu)^2}{\sigma^5\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}=\frac{1}{\sigma^5\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}(\sigma^2-(x - \mu)^2)\).
Set \(y^{\prime\prime}=0\), then \(\sigma^2-(x - \mu)^2 = 0\).
Solving \((x - \mu)^2=\sigma^2\), we get \(x=\mu-\sigma\) and \(x=\mu+\sigma\).

Answer:

B. The points are \(x = \mu-\sigma\) and \(x=\mu+\sigma\)