Question

Suppose that a random variable x can take on integer values from 0 to 5 and its pdf is defined as follows:P(x)=11-2x/36 Calculate the expected values of x

Answer 1

Given the pdf defined:

$P(X=x)=\frac{11-2x}{36}$

The formula to find the expected value of x for a discrete distribution is

$E(X)=\sum ^{\infty}_{n\mathop=-\infty}xP(X=x)$

Here, x ranges from 0 to 5. Find E(X).

$\begin{gathered} E(X)=\sum ^5_{n\mathop=0}xP(X=x) \\ =0\cdot P(x=0)+1\cdot P(x=1)+2\cdot P(x=2)+3\cdot P(x=3)+4.P(x=4)+5\cdot P(x=5)P( \\ =0+1\cdot\frac{11-2\cdot1}{36}+2\cdot\frac{11-2\cdot2}{36}+3\cdot\frac{11-2\cdot3}{36}+4\cdot\frac{11-2\cdot4}{36}+5\cdot\frac{11-2\cdot5}{36} \\ =\frac{9}{36}+\frac{14}{36}+\frac{15}{36}+\frac{12}{36}+\frac{5}{36} \\ =\frac{55}{36} \end{gathered}$

which is the expected value of x.