A normal distribution in statistics is modeled by the function $N(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}$ determine the limit of N(x)as $x \rightarrow$ - $\infty$
A)- $\infty$
B)- $\frac{1}{2 \pi}$
C)0
D) $\frac{1}{2 \pi}$
E) $\infty$

Question

Quizplus · Accepted Answer

As $x$ approaches $-\infty$, the exponent $-x^2/2$ in the function $N(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}$ becomes very large and negative, causing $e^{-x^{2} / 2}$ to approach 0. Therefore, $N(x)$ approaches 0.

A Normal Distribution in Statistics Is Modeled by the Function N(x)=12πe−x2/2N(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}N(x)=2π​1​e−x2/2

A Normal Distribution in Statistics Is Modeled by the Function $N(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}$