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## Homework Statement

Let X ~ norm(5,10). Find P(X>10).

## Homework Equations

f(x) = [itex]\frac{1}{δ\sqrt{2π}} e^{-\frac{(x-μ)^2}{2δ^2}}[/itex]

F(x) = P(X<x) = [itex]\int_{-∞}^x f(u) du[/itex]

## The Attempt at a Solution

P(X>10) = 1 - P(X<10)

P(X<10) = [itex]\int_{-∞}^{10} \frac{1}{δ\sqrt{2π}} e^{-\frac{(x-μ)^2}{2δ^2}} dx[/itex]

= [itex]\frac{1}{\sqrt{2π}} \int_{-∞}^{10} \frac{1}{δ} e^{-\frac{(x-μ)^2}{2δ^2}} dx[/itex]

Let [itex]-\frac{(x-μ)^2}{2δ^2} = -\frac{y^2}{2}[/itex], then

[itex]y = \frac{x-μ}{δ}[/itex], Since [itex]μ = 5, δ = 10 → y = 0.5[/itex], when [itex]x = 10[/itex]

[itex]y = \frac{x-μ}{δ}[/itex], Since [itex]μ = 5, δ = 10 → y = -∞[/itex], when [itex]x = -∞[/itex]

[itex]x = δy + μ[/itex], So [itex]dx = δ dy[/itex].

So P(X<10) = [itex]\frac{1}{\sqrt{2π}} \int_{-∞}^{0.5} e^{-\frac{y^2}{2}} dy[/itex]

= [itex]\frac{1}{\sqrt{2π}} (\int_{-∞}^{0.5} e^{-\frac{y^2}{2}} dy \int_{-∞}^{0.5} e^{-\frac{x^2}{2}} dx )^{0.5}[/itex]

= [itex]\frac{1}{\sqrt{2π}} (\int_{0}^{2π} \int_0^{0.5}re^{-\frac{r^2}{2}} drdθ)^{0.5}[/itex]

= [itex]\frac{1}{\sqrt{2π}} (\int_{0}^{2π} [-e^{-\frac{r^2}{2}}]\stackrel{0}{0.5} dθ)^{0.5}[/itex]

= [itex]\frac{1}{\sqrt{2π}} (\int_{0}^{2π} 1-e^{-\frac{1}{8}}dθ)^{0.5}[/itex]

= [itex]\frac{1}{\sqrt{2π}} (2π (1-e^{-\frac{1}{8}})^{0.5})[/itex]

= [itex](1-e^{-\frac{1}{8}})^{0.5}[/itex]

So [itex]P(X>10) = 1 - (1-e^{-\frac{1}{8}})^{0.5} = 0.6572...[/itex], but it should be 0.3085.... What went wrong?

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