Provide an appropriate response.
-The standard normal probability density function is defined by $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } / 2 }$.
(a) Show that $\int _ { 0 } ^ { \infty } \frac { 1 } { \sqrt { 2 \pi } } x \mathrm { e } ^ { - x ^ { 2 } / 2 } \mathrm { dx } = \frac { 1 } { \sqrt { 2 \pi } }$.
(b) Use the result in part (a) to show that the standard normal probability density function has mean 0 .

Question

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The Answer of Provide an appropriate response.
-The standard normal probability density function is defined by $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } / 2 }$.
(a) Show that $\int _ { 0 } ^ { \infty } \frac { 1 } { \sqrt { 2 \pi } } x \mathrm { e } ^ { - x ^ { 2 } / 2 } \mathrm { dx } = \frac { 1 } { \sqrt { 2 \pi } }$.
(b) Use the result in part (a) to show that the standard normal probability density function has mean 0 .

Provide an Appropriate Response f(x)=12πe−x2/2\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } / 2 }f(x)=2π​1​e−x2/2

Provide an Appropriate Response $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } / 2 }$