Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive constant. 1. find c. 2. compute p(x > 2). 3. compute e[x] and var(x). 4. let y = x2+1, find the pmf of y: 5. compute

Looks like the PMF is supposed to be



which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:



Next,









If , then , where we take the positive root because we know can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that can take on the values , , and . At these values of , we would have the same probability as we did for the respective value of . That is,



Part (5) is incomplete, so I'll stop here.

Aquadratic function is one of the form f(x)=ax2 + bx + c l, where a,b, and c are numbers with are not equal to zero.

They are somewhat similar