Find the mean, variance, and standard deviation of the normal density function $f ( x ) = \frac { 1 } { 2 \sqrt { 2 \pi } } e ^ { - ( x -- 1 ) ^ { 2 }/8}$ over $( - \infty , \infty )$ . Do not use integration.
A) $\begin{array} { l l }
\text { expected value(mean): } & - 1 \
\text { variance: } & 4 \
\text { standard deviation: } & 2
\end{array}$
B) $\begin{array} { l l }
\text { expected value (mean) } ) : & 4 \
\text { variance: } & - 1 \
\text { standard dev iation: } & 2
\end{array}$
C) $\begin{array} { l l }
\text { expected value(mean } ) : & 4 \
\text { variance: } & 2 \
\text { standard dev iation: } & - 1
\end{array}$
D) $\begin{array} { l l }
\text { expected value (mean): } & 2 \
\text { variance: } & - 1 \
\text { standard deviation: } & 4
\end{array}$
E) $\begin{array} { l l }
\text { expected value (mean): } & - 1 \
\text { variance: } & 2 \
\text { standard deviation: } & 4
\end{array}$

Question

Find the mean, variance, and standard deviation of the normal density function $f ( x ) = \frac { 1 } { 2 \sqrt { 2 \pi } } e ^ { - ( x -- 1 ) ^ { 2 }/8}$ over $( - \infty , \infty )$ . Do not use integration.
A) $\begin{array} { l l } 
\text { expected value(mean): } & - 1 \
\text { variance: } & 4 \
\text { standard deviation: } & 2
\end{array}$
B) $\begin{array} { l l } 
\text { expected value (mean) } ) : & 4 \
\text { variance: } & - 1 \
\text { standard dev iation: } & 2
\end{array}$
C) $\begin{array} { l l } 
\text { expected value(mean } ) : & 4 \
\text { variance: } & 2 \
\text { standard dev iation: } & - 1
\end{array}$
D) $\begin{array} { l l } 
\text { expected value (mean): } & 2 \
\text { variance: } & - 1 \
\text { standard deviation: } & 4
\end{array}$
E) $\begin{array} { l l } 
\text { expected value (mean): } & - 1 \
\text { variance: } & 2 \
\text { standard deviation: } & 4
\end{array}$

Quizplus · Accepted Answer

The correct mean is -1, as indicated by the term $(x - (-1))^2$ in the exponent of the density function, which shows the mean is -1. The variance is given by the coefficient in the denominator of the exponent, which is 8, but since the standard form is $\frac{(x - \mu)^2}{2\sigma^2}$, the variance $\sigma^2$ is 4. The standard deviation is the square root of the variance, which is 2.

Find the Mean, Variance, and Standard Deviation of the Normal f(x)=122πe−(x−−1)2/8f ( x ) = \frac { 1 } { 2 \sqrt { 2 \pi } } e ^ { - ( x -- 1 ) ^ { 2 }/8}f(x)=22π​1​e−(x−−1)2/8

Find the Mean, Variance, and Standard Deviation of the Normal $f ( x ) = \frac { 1 } { 2 \sqrt { 2 \pi } } e ^ { - ( x -- 1 ) ^ { 2 }/8}$