The discrete random variable W has probability density function (p.d.f.) given by:
P(W=w) = c(5-w) for w= 0, 1, 2, 3
Find c
I honestly have no idea how to start Any help is appreciated.
Thanks.
the probabilities must add up to 1. So,
c(5-0)+c(5-1)+c(5-2)+c(5-3) = 1
5c+4c+3c+2c = 1
c = 1/14
well, if that is the whole distribution, they have to add up to 1
5 c +4 c + 3 c + 2 c = 1
Oh wow it was a lot easier than I thought...
Thanks so much!
You are welcome.
To find the value of c in the given problem, we need to use the fact that the sum of the probabilities for all possible values of W should be equal to 1.
In this case, we have the probabilities P(W = w) for w = 0, 1, 2, and 3, given by P(W = 0) = c(5-0), P(W = 1) = c(5-1), P(W = 2) = c(5-2), and P(W = 3) = c(5-3).
To find c, we sum up these probabilities and set the sum equal to 1:
P(W = 0) + P(W = 1) + P(W = 2) + P(W = 3) = 1
Substituting the given probabilities, we have:
c(5-0) + c(5-1) + c(5-2) + c(5-3) = 1
Simplifying the equation, we get:
c(5) + c(4) + c(3) + c(2) = 1
Now, let's factor out c from each term on the left side:
c(5 + 4 + 3 + 2) = 1
c(14) = 1
Finally, dividing both sides of the equation by 14:
c = 1/14
Therefore, the value of c that satisfies the given conditions is 1/14.