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The question is from a past paper with no solutions, so I am trying to complete them and confirm with you guys if this is correct.

a) Findingk, I solved for

$\displaystyle 1 = k \left[ \int_{-1}^{0} (x+2)\ dx + \int_{0}^{1} (1-x^2)\ dx\ \right]$

for which I get $\displaystyle k=\frac{6}{13}$

b) the CDF, I just took the integrals of all the terms so it should look like

0 if $\displaystyle x < -1$

$\displaystyle \frac{x^2}{2} + 2x$ if $\displaystyle -1 \leq x \leq 0$

$\displaystyle x - \frac{x^3}{3}$ if $\displaystyle 0 \leq x \leq 1$

0 if $\displaystyle x > 1$

c) Expected value of X

I want to confirm this before calculating the variance, because I am getting a negative mean of $\displaystyle -\frac{5}{26}$

I am attempting to calculate it as follows:

$\displaystyle E[X] = \frac{6}{13} \left[ \int_{-1}^{0} x(x+2)\ dx + \int_{0}^{1} x(1-x^2)\ dx \right]$

e) Find the probability that $\displaystyle X>0$

It is greater than 0 if it is between [0,1], so I've taken the integral:

$\displaystyle \frac{6}{13} \int_{0}^{1} (1-x^2)\ dx = \frac{4}{13} $

Apologies the question is long, but they are related.