Random variables are those who have their values dependent on the outcomes of a random experiment. Random variables can be discrete or continuous. The discrete random variable takes only discrete values while the continuous ones may take infinite random values in an interval. Probability distribution shows the probability for different discrete values. Cumulative probability is the probability of getting a value of random variable less than or equal to a certain value. 

Word Problems

Let us have a look on few examples.
Problem 1: 

A couple has three kids. What will be the values which can be attained by the random variable representing number of daughters the couple is having?

Solution: 

Let the random variable be $X$.

$X$ = Number of daughters the couple is having

The discrete values for $X$ will be $\{0,\ 1,\ 2,\ 3\}$ as out of three kids there can be none, one, two or all three daughters.
Problem 2:

From the given random variables, find out which ones are discrete and which are continuous.

a) Speed of train

b) Number of students getting A grade

c) Height of men in Alaska

d) Error in measurement

Solution: 

The random variables can be classified as given here,

a) Speed of train will have continuous values and hence, it is a continuous random variable.

b) Number of students will have countable values and hence, it is a discrete random variable.

c) Height of men will have continuous values and hence, it is a continuous random variable.

d) Error in measurement will have continuous values and hence, it is a continuous random variable.
Problem 3:

Construct cumulative distributive function for the given probability distribution.

 X 0
 1  2 3
P(X=k)  0.3  0.2 0.4
 0.1

Solution:

The cumulative distribution for $P(X\ \leq\ k)$ is as in the given table.

 X 0  1  2  3
 P(X=k)  0.3  0.2  0.4  0.1
 P(X$\leq $k  0.3  0.5  0.9  1
Problem 4:

Let $X$ be a random variable defining number of students getting A grade. Find the expected value of $X$ from the given table.

 X  0  1  2  3
 P(X)  0.2 0.1
0.4
0.3

Solution: The expected value of X is given by the formula $E(X)$ = $\sum x_ip_i$.

The given data can be calculated as,

 X  P(X)  $x_ip_i$
 0  0.2  0
 1  0.1  0.1
 2  0.4  0.8
 3  0.3  0.9

 The expected value will be $E(X)$ = $\sum x_ip_i$ = $1.8$.