Expected value, as the name suggest, it is the return value expected after an action. If probability of getting aresult is P, and number of experiments is n then the expected value for binomial variable can be calculated as, 
$E(X)\ =\ P\ \times\ n$. 

Expected value of a discrete random variable is the mean of the list of values given. If probability distribution function $f(x)$ is given, then find the value of $f(x)$ for different values of x and check where the sum is converging to.

Example 1: 

If marks of five students is given to be 65, 76, 88, 34, and 90, then find expected value of mark for a random student.


As discrete values are given, the expected value is the mean of all the values given.
Expected value, $E(X)$ = $\frac{65 + 76 + 88 + 34 + 90}{5}$ = $\frac{353}{5}$ = $70.6$

Expected Value Word Problems

Problem 1:

If height of four women is given to be 150 cm, 165 cm, 135 cm and 170 cm, then what will be the expected value of the height of a randomly chosen women?


The expected value is the mean of all the values.

$E(X)$ = $\frac{150+165+135+170}{4}$ = $\frac{620}{4}$ = $155$

The expected value is found to be 155 cm.
Problem 2:

Find the expected value of X from the given probability distribution table.

 1 2
 0.8  0.6 0.2 


The expected value of X, E(X) = $\sum X\times P(X)$.

$E(X)$ = $1\ \times\ 0.8\ +\ 2\ \times\ 0.6\ +\ 3\ \times\ 0.2$
           = $0.8\ +\ 1.2\ +\ 0.6\ =\ 2.4$
Problem 3:

There are four balls in a bag, red, black, green and blue. There is equal probability of getting any colored ball. What is the expected value of getting a green ball out of 20 experiments with replacement?


Probability of getting a green ball, $P$ = $\frac{1}{4}$ = $0.25$

$Expected\ Value$ = $Number\ of\ Experiments\ \times\ Probability$

$E(X)\ =\ 20\ \times\ 0.25\ =\ 5$
Problem 4:

The probability density function for an event is given as, $f(x)$ = $\frac{1}{2x}$, then find the expected value.


Putting the different values of x to get f(x),

 1  2  3  4
 $\frac{1}{2}$  $\frac{1}{4}$  $\frac{1}{6}$  $\frac{1}{8}$

 Adding these values,

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{6}$ + $\frac{1}{8}$ = $\frac{27}{24}$ = $1.125$

Hence, the expected value will converge to $1.2.$