 # help with statistics

Hey Lesley! Welcome.

Since I have limited time to answer people, I kindly ask that you don’t post a whole bunch of questions into one thread. Let’s try one question next time, as in, post a), b), c) or d), but not all of them at once. Also, it’d be more productive if you ask a specific question, rather than post without letting me know what you’re not understanding.

That said, I’ll show you how to do a).

(i)
Let’s start with i). We want to show that f is a PDF and to find the value of C.

To show that f is a PDF, we have to show that f satisfies two properties: f(x) \ge 0 and \int_{\infty}^{\infty} f(x) = 1. Since 4x - 2x^2 = 2x(2-x), and since each of these factors is positive over the interval (0,2), we know that the whole expression is positive over the same interval, and so we have shown the first part. For the second part, solve the integral:

\begin{align*} \int_{-\infty}^{\infty} C(4x - 2x^2) ~\text{d}x &= C\int_{-\infty}^{\infty} (4x - 2x^2) ~\text{d}x \\ &= C\int_{0}^{2} (4x - 2x^2) ~\text{d}x \\ &= C\left[ 2x^2 - \frac{2}{3}x^3 \right]_0^{2} \\ &= C\left[ 2(2)^2 - \frac{2}{3}(2)^3\right] \\ &=C\frac{8}{3} \end{align*}\tag*{}

In order for f to be a PDF, this integral must equal 1, to it must be the case that C=\frac{3}{8}.

(ii)

\begin{align*} p(x>1) &= \int_{1}^{\infty} \frac{3}{8}(4x - 2x^2) ~\text{d}x \\ &= \int_{1}^{2} \frac{3}{8}(4x - 2x^2) ~\text{d}x \\ &= \frac{3}{8}\left[ 2x^2 - \frac{2}{3}x^3 \right]_1^{2} \\ &= \frac{1}{2} \end{align*}\tag*{}

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