Step 1

Here, \(x\sim=N(M,\sigma^2)\)

To find \( \int_{0}^{+oo} p(x)dx, 10\) here \(\int_{}^{}(x)=\frac{1}{\sigma\sqrt{2K}}e^(x-\mu)^2/2\sigma^2,x\in R\)

\( \int_{}^{}(x)\)is the PDF of x

\(\displaystyle\Rightarrow{\int_{{{0}}}^{{+\infty}}}{\int_{{\lbrace}}^{{\lbrace}}}{\left({x}\right)}{\left.{d}{x}\right.}={P}{\left[{x}{>}{0}\right]}\)

\(\displaystyle{x}\sim{N}{\left({M},\sigma^{{2}}\right)}\Rightarrow{\frac{{{x}-{M}}}{{\sigma}}}\sim{N}{\left({0},{1}\right)}\)

\(\displaystyle{F}{\quad\text{or}\quad}{M}={0},\sigma={5}{M}={\frac{{{x}}}{{{5}}}}\sim{N}{\left({0},{1}\right)}\)

\(\displaystyle{T}{h}{e}{n},{P}{\left[{x}{>}{0}\right]}={P}{\left[{\frac{{{x}}}{{{5}}}}{>}{0}\right]}+{P}{\left[{y}{>}{0}\right]}\)

\(\displaystyle={1}-{P}{\left[{y}\leq{0}\right]}={1}-\phi{\left({0}\right)}\)

\(\displaystyle={1}-{0}\cdot{5}={0}\cdot{5}\) Step 2 Hence,the required value is 0.5.