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A continuous random variable $X$ has cumulative distribution function

$$f(x) = \begin{cases} 0 &\textrm{ when } x<0,\\ x^2(3-2x) &\textrm{ when } 0\leq x \leq 1,\\ 1 &\textrm{ when } x>1. \end{cases}$$

(i) Calculate $\mathrm{Pr}\left(\frac13 < X < \frac23\right)$.

For this I got $\mathrm{Pr}(a<x<b) = \mathrm{Pr}(x\leq b) - \mathrm{Pr}(x\leq a)$ hence I got $20/27 - 7/27 = 13/27$.

(ii) Find the probability density function of $X$.

For this I got the derivative to equal $0$ when $x<0$ and $x>1$ and to equal $6x(1-x)$ for $0 \leq x \leq 1$

(iii) The symmetry of the probability density function on the interval $[0, 1]$ implies that that $E(X) = \frac12$. Find $\mathrm{var}(X)$.

Not really sure how to answer this?? Also could someone confirm if my other two answers are correct Thankyou :)

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Hint: $\text{var}(X) = E[X^2]-(E[X])^2$ where the value of $E[X]$ is given to you. Can you find $E[X^2]$? –  Dilip Sarwate Dec 10 '12 at 19:36
i'm struggling to work out $E[X^2]$ –  jill Dec 10 '12 at 20:02
i know $E[X^2] = \mew^2 + \sig^2$ and that $E[X] = \mew$ –  jill Dec 10 '12 at 20:10
@AndréNicolas heyaaa don't know if this is your area of expertise but any help would be much appreciated :) –  jill Dec 10 '12 at 20:39
Further hint: $E[X^2] = \int_{-\infty}^\infty x^2 (\text{density function of}~ X)\,\mathrm dx$. Your mew is given to you, and your $\text{sig}^2$ is what you are asked to find when you are asked to find the variance of $X$. So, finding $\text{sig}^2$ first, then computing $E[X^2] = \text{mew}^2+\text{sig}^2$, and finally finding $\text{var}(X)=\text{sig}^2$ as $E[X^2]-\text{mew}^2$ is kind of silly, don't you think? –  Dilip Sarwate Dec 10 '12 at 22:15
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