# Use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the functionp(x)=\frac{1}{\sqrt{2\pi^\sigma}}e-(x-\mu)^2/(2\sigma^2)

This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function
$$\displaystyle{p}{\left({x}\right)}={\frac{{{1}}}{{\sqrt{{{2}\pi^{\sigma}}}}}}{e}-\frac{{\left({x}-\mu\right)}^{{2}}}{{{2}\sigma^{{2}}}}$$
where $$\pi$$ = 3,14159265 ... and $$\sigma$$ and $$\mu$$ are constants called the standard deviation and the mean, respectively. Its graph (for $$\sigma=1$$ and $$\mu=2)$$ is shown in the figure.
With $$\displaystyle\sigma={\color{red}{{5}}}$$ and $$\mu=0$$, approximate $$\displaystyle{\int_{{{0}}}^{{+\infty}}}{p}{\left({x}\right)}{\left.{d}{x}\right.}$$.(Round your answer to four decimal places.)

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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.