density function of a dealers profit

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?

Re: density function of a dealers profit

"Usually" is a bit too general (you may be thinking of Cheby when dealing with a standard normal). In any case all you're asked to do is verify that the probability that any r.v. falls outside your specified limits (on the left or the right) is simply 1/4 in total. This amounts to integrating from -1 to -7/9 and, 7/9 to 1, and summing the result (I'll leave the whys to you).