Discrete and random variable....i need help...?

Question

A discrete random variable X takes integer values between 0 and 6 inclusive with probabilites given by P(X=x)= 1/4- (3-x)/16. find the expectation and variance of X

Answer 1

You are told that the values that X can be are 0, 1, 2, 3, 4, 5 and 6
You can then calculate the probability of X taking each of those values by, one by one, plugging that value in (as small x) the equation given
The rest should follow

Answer 2

As given, the probabilities for x = 0, 1, 2, 3, 4, 5, 6 work out to be 1/16, 2/16, 3/16, 4/16, 5/16, 6/16, 7/16 and they do not add up to1 and therefore there is something wrong in the question. This is not a valid probability distribution.
There is no point in trying to work out expectation and variance until you have the probability distribution correct.

When you do have it correctly defined and have worked out individual probabilities then apply
E(X) = x*P(x) summed over all x (Let this equal m)
Var(X) = (x - m)^2*P(X) summed over all x.
However Var(X) = (sum x^2*P(X)) - m^2 is often easier.

Answer 3

A discrete random variable is one which will handle in basic terms a countable form of different values which includes 0,a million,2,3,4,........ Discrete random variables are many times (yet not inevitably) counts. If a random variable can take in basic terms a finite form of different values, then it might desire to be discrete. Examples of discrete random variables incorporate the form of childrens in a kinfolk, the Friday night attendance at a cinema, the form of sufferers in a doctor's surgical treatment, the form of defective mild bulbs in a container of ten. so specific, you does not say sixty two a million/2 beats in step with minute... you may say sixty two bpm or sixty two bpm My source is Yale college Stat branch and that i hear they're especially clever up there.