# using the probability density function to find the distribution function and determine certain x values

would anyone be able to help me with this question?

Consider a random variable $X$ with probability density function

$$f(x) = xe^{−x^2/2}, x\geq 0.$$

Determine x such that

$P(X \lt x) = 0.5$, $P(X \leq x) = 0.5$, and $P(X \gt x) = 0.95.$

What I've tried to do is integrate $f(x)$ to get $F(X)$ and then substitue in the different values of $F(X)$. However I haven't been getting real numbers as answers.

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The error is a fairly common one, and easy to fix.

First find $F(x)$ (it is $x$, not your $X$). We have, for $x \ge 0$, $$F(x)=\int_0^x te^{-t^2/2}dt$$

Integrate. One antiderivative is $-e^{-t^2/2}$. This is $-e^{-x^2/2}$ at $x$ and $-1$ at $0$, so (after tiny simplification) $F(x)=1-e^{-x^2/2}$ for $x \ge 0$.

Your mistake was to assume without calculation that the value of our antiderivative is $0$ at $0$. (When integrating polynomials, we get used to kind of forgetting about the value at $0$.)

The rest should be straightforward.

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You want to find $x_1$ and $x_2$ such that $$\int_{0}^{x_1} xe^{-x^2/2} \ dx = 0.5$$ and $$\int_{0}^{x_2} xe^{-x^2/2} \ dx = 0.05$$ where the first one is for parts 1 and 2, and the second one is for the third part (i.e $P(X > x) = 0.95 \implies P(X \leq x) = 0.05$).

For parts 1 and 2 I get $x_1 \approx 1.7741$. For the third part I get $x_2 \approx 0.320$.

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