A dealers profit in units of 5,000 on a new auto there is a random variable X having density function

$\displaystyle f(x) = 1/2(1-x) for -1\le x\le1$

a) find the variance in the dealers profit

b) Demonstrate that chebyshevs inequality holds for k=2 with the density function above

c) what is the probobility that the profit exceeds $500

For part a i got

$\displaystyle E[X]= -\frac{1}{3}$

which i know is correct.

$\displaystyle E(X^2) = \frac{1}{3}$

which i know is correct.

Therefore,

$\displaystyle Var[x] = E[X^2] - E[X]^2 = \frac{2}{9}$

for part b chebyshevs inequality

Chebyshev's inequality is usually written as $\displaystyle Pr( |X-M| > ks) < 1/k^2.$

What we know:

$\displaystyle E[X] = M = -1/3$

$\displaystyle Sigma = s = \frac{2}{9}$

$\displaystyle k = 2$

adding them into the chebyshevs we have

$\displaystyle Pr( |X-(-1/3)| > 2* 2/9) < 1/k^2.$

$\displaystyle Pr( |X+1/3)| > 4/9) < 1/4$

I dont know what to do from here. Would i move the 1/3 to the right side of the inequality?

then use that as a limit in an integral?

for part c

$\displaystyle PR(X> \frac{1}{10})$

therefore use lower limit 1/10 and upper limit 1 in an integral?