The PMF of the random variable Y satisfies pY(−1)=1/6 , pY(2)=2/6 , pY(5)=3/6 , and pY(y)=0 for all other values y . The expected value of Y is:
E[Y]=
i got the right answer its 3
not 3 frequency cant be negative
Well, let's calculate the expected value of Y, shall we?
First, we need to multiply each possible value of Y by its corresponding probability, and then sum them all up.
E[Y] = (-1)(1/6) + (2)(2/6) + (5)(3/6)
Simplifying this expression, we get:
E[Y] = -1/6 + 4/6 + 15/6
Combining the fractions, we have:
E[Y] = 18/6
Now, reducing the fraction, we get:
E[Y] = 3
So, the expected value of Y is 3. Now that's an expected to get excited about!
To find the expected value E[Y] of a random variable Y, you need to multiply each possible value of Y by its corresponding probability and then sum them up.
In this case, the possible values of Y are -1, 2, and 5, with their respective probabilities being 1/6, 2/6, and 3/6.
So, to calculate E[Y], you can use the formula:
E[Y] = (-1)*(1/6) + 2*(2/6) + 5*(3/6)
Simplifying the expression:
E[Y] = -1/6 + 4/6 + 15/6
Adding the fractions:
E[Y] = (18 - 1 + 4)/6
E[Y] = 21/6
Simplifying the fraction:
E[Y] = 7/2
Therefore, the expected value of Y is 7/2.