- #1
Catchfire
- 30
- 0
Homework Statement
Let X ~ norm(5,10). Find P(X>10).
Homework Equations
f(x) = \frac{1}{δ\sqrt{2π}} e^{-\frac{(x-μ)^2}{2δ^2}}
F(x) = P(X<x) = \int_{-∞}^x f(u) du
The Attempt at a Solution
P(X>10) = 1 - P(X<10)
P(X<10) = \int_{-∞}^{10} \frac{1}{δ\sqrt{2π}} e^{-\frac{(x-μ)^2}{2δ^2}} dx
= \frac{1}{\sqrt{2π}} \int_{-∞}^{10} \frac{1}{δ} e^{-\frac{(x-μ)^2}{2δ^2}} dx
Let -\frac{(x-μ)^2}{2δ^2} = -\frac{y^2}{2}, then
y = \frac{x-μ}{δ}, Since μ = 5, δ = 10 → y = 0.5, when x = 10
y = \frac{x-μ}{δ}, Since μ = 5, δ = 10 → y = -∞, when x = -∞
x = δy + μ, So dx = δ dy.
So P(X<10) = \frac{1}{\sqrt{2π}} \int_{-∞}^{0.5} e^{-\frac{y^2}{2}} dy
= \frac{1}{\sqrt{2π}} (\int_{-∞}^{0.5} e^{-\frac{y^2}{2}} dy \int_{-∞}^{0.5} e^{-\frac{x^2}{2}} dx )^{0.5}
= \frac{1}{\sqrt{2π}} (\int_{0}^{2π} \int_0^{0.5}re^{-\frac{r^2}{2}} drdθ)^{0.5}
= \frac{1}{\sqrt{2π}} (\int_{0}^{2π} [-e^{-\frac{r^2}{2}}]\stackrel{0}{0.5} dθ)^{0.5}
= \frac{1}{\sqrt{2π}} (\int_{0}^{2π} 1-e^{-\frac{1}{8}}dθ)^{0.5}
= \frac{1}{\sqrt{2π}} (2π (1-e^{-\frac{1}{8}})^{0.5})
= (1-e^{-\frac{1}{8}})^{0.5}
So P(X>10) = 1 - (1-e^{-\frac{1}{8}})^{0.5} = 0.6572..., but it should be 0.3085... What went wrong?
Last edited:
