Question

A probability distribution for a random variable Y is given by P(Y=y)=cy, for y=1,2,4,5 and c is a constant. Find E(Y), the expected value of Y. Give your answer as a fraction in its simplest form.

Answer 1

Answer:

$c(1+2+4+5) =1$

$c =\frac{1}{12}$

Now we can find the expected value with this formula:

$E(Y) =\sum_{i=1}^n Y_i P(Y_i) =\sum_{i=1}^n cy_i *y_i = cy^2_i$

And replacing we got:

$E(Y) = \frac{1}{12} (1^2 + 2^2 + 4^2 +5^2) = \frac{46}{12}= \frac{23}{6}$

Step-by-step explanation:

For this case we know the following probability mass function given:

$P(y) = cy , y= 1,2,4,5$

For this case we need to satisfy the following condition in order to have a probability distribution function:

$\sum_{i=1}^n P(y_i) =1$

And we have this:

$c(1+2+4+5) =1$

$c =\frac{1}{12}$

Now we can find the expected value with this formula:

$E(Y) =\sum_{i=1}^n Y_i P(Y_i) =\sum_{i=1}^n cy_i *y_i = cy^2_i$

And replacing we got:

$E(Y) = \frac{1}{12} (1^2 + 2^2 + 4^2 +5^2) = \frac{46}{12}= \frac{23}{6}$