5.4.1 The Probability Distribution Table

In this topic we will learn how to:

draw a probability distribution table relating to a given situation involving a discrete random variable $X$ , and calculate $E(X)$ and $Var(X)$

A random variable is one that can take multiple values and in some cases these values may be defined by a function. Random variables are denoted by a capital letter, for example, $X$ . A discrete random variable only takes exact values, for example, $1$ , $2$ , $3$ , and so on. The sum of the probabilities of all the values governed by a particular random variable adds up to $1$ . This can be denoted by,
$P(X = x) = 1$ Where $X$ represents the random variable and $x$ represents all the values that the random variable can take.
This may also be written as,
$p = 1$ Where $p$ represents $P(X = x)$ .

If we know the probabilities of the values that the random variable can take, we can draw a probability distribution table.
Let’s look at an example.

Example 1
The random variable $X$ , takes values $1$ , $2$ , $3$ and $4$ . It is given that $P(X = 1) = 0.1$ , $P(X = 2) = 0.4$ , $P(X = 3) = 0.4$ and $P(X = 4) = 0.1$ . Draw the probability distribution table.

The probability distribution table for the problem above would look like this,

$x$	$1$	$2$	$3$	$4$
$P(X = x)$	$0.1$	$0.4$	$0.4$	$0.1$

Using the probability distribution table we can also find the $E(X)$ and $Var(X)$ . The formula for mean is,
$E(X) = \Sigma xp$ The formula for variance is,
$Var(X) = \Sigma x^{2}p - [E(X)]^{2}$ Using the example above, let’s find $E(X)$ and $Var(X)$ ,
$E(X) = \Sigma xp$ $E(X) = 1(0.1) + 2(0.4) + 3(0.4) + 4(0.1)$ $E(X) = 2.5$ $Var(X) = \Sigma x^{2}p - [E(X)]^{2}$ $\Sigma x^{2}p = 1^{2}(0.1) + 2^{2}(0.4) + 3^{2}(0.4) + 4^{2}(0.1)$ $\Sigma x^{2}p = 6.9$ $Var(X) = 6.9 - (2.5)^{2}$ $Var(X) = 0.65$

Let’s look at some past paper questions.

1. The random variable $X$ takes the values $1$ , $2$ , $3$ , $4$ only. The probability that $x$ takes the value $x$ is $kx(5 - x)$ , where $k$ is a constant. (9709/52/F/M/21 number 4)

(a) Draw up the probability distribution table for $X$ , in terms of $k$ .

We are given that,
$P(X = x) = kx(5 - x)$ We can use that function to find the probabilities for the values that $x$ takes. Let’s start with $1$ ,
$P(X = x) = kx(5 - x)$ $P(X = 1) = k(1)(5 - 1)$ $P(X = 1) = 4k$ Let’s evaluate $2$ ,
$P(X = x) = kx(5 - x)$ $P(X = 2) = k(2)(5 - 2)$ $P(X = 2) = 6k$ Let’s evaluate $3$ ,
$P(X = x) = kx(5 - x)$ $P(X = 3) = k(3)(5 - 3)$ $P(X = 3) = 6k$ Let’s evaluate $4$ ,
$P(X = x) = kx(5 - x)$ $P(X = 4) = k(4)(5 - 4)$ $P(X = 4) = 4k$ Now that we have all the probabilities in terms of $k$ , we can now draw up a probability distribution table,

$x$	$1$	$2$	$3$	$4$
$P(X = x)$	$4k$	$6k$	$6k$	$4k$

Therefore, the final answer is,

$x$	$1$	$2$	$3$	$4$
$P(X = x)$	$4k$	$6k$	$6k$	$4k$

(b) Show that $Var(X) = 1.05$ .

The formula for variance under discrete conditions is,
$Var(X) = \Sigma x^{2}p - [E(X)]^{2}$ To be able to use this formula we need to find $\Sigma x^{2}p$ and the mean. To do that, we need to first evaluate $k$ , we will use the idea that,
$P(X = x) = 1$ $P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$ Substitute in the values in terms of $k$ from the probability distribution table,
$4k + 6k + 6k + 4k = 1$ Simplify and solve for $k$ ,
$20k = 1$ $k = \frac{1}{20}$ We can now draw up a new probability distribution table in which the $k$ has been substituted,

$x$	$1$	$2$	$3$	$4$
$P(X = x)$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{6}{20}$	$\frac{4}{20}$

We can now find the mean,
$E(X) = \Sigma xp$ $E(X) = 1\left(\frac{4}{20}\right) + 2\left(\frac{6}{20}\right) + 3\left(\frac{6}{20}\right) + 4\left(\frac{4}{20}\right)$ $E(X) = 2.5$ Note: If the distribution is symmetrical, then the mean is the midpoint of the $x$ values.

Now let’s find $\Sigma x^{2}p$ ,
$\Sigma x^{2}p = 1^{2}\left(\frac{4}{20}\right) + 2^{2}\left(\frac{6}{20}\right) + 3^{2}\left(\frac{6}{20}\right) + 4^{2}\left(\frac{4}{20}\right)$ $\Sigma x^{2}p = 7.3$ Now let’s substitute back into the formula for variance,
$Var(X) = \Sigma x^{2}p - [E(X)]^{2}$ $Var(X) = 7.3 - (2.5)^{2}$ $Var(X) = 7.3 - 6.25$ $Var(X) = 1.05$ Therefore, we have proved that,
$Var(X) = 1.05$ 2. A fair spinner has sides numbered $1$ , $2$ , $2$ . Another fair spinner has sides numbered $-2$ , $0$ , $1$ . Each spinner is spun. The number on the side on which a spinner comes to rest is noted. The random variable $X$ is the sum of the numbers for the two spinners. (9709/52/M/J/21 number 4)

(a) Draw up the probability distribution table for $X$ .

Since there are two spinners, we can draw a possibility space diagram to show all possible outcomes. Label the $y$ -axis with, spinner $1$ and the $x$ -axis with, spinner $2$ . Inside the diagram add up the sum of the numbers for the two spinners,

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From the diagram we can tell that,
$P(X = -1) = \frac{1}{9}$ $P(X = 0) = \frac{2}{9}$ $P(X = 1) = \frac{1}{9}$ $P(X = 2) = \frac{3}{9}$ $P(X = 3) = \frac{2}{9}$ Now that we have all the probabilities, we can now draw a probability distribution table,

$x$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{9}$	$\frac{3}{9}$	$\frac{2}{9}$

Therefore, the final answer is,

$x$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{1}{9}$	$\frac{3}{9}$	$\frac{2}{9}$

(b) Find $E(X)$ and $Var(X)$ .

To find the mean, we use the formula,
$E(X) = \Sigma xp$ $E(X) = -1\left(\frac{1}{9}\right) + 0\left(\frac{2}{9}\right) + 1\left(\frac{1}{9}\right) + 2\left(\frac{3}{9}\right) + 3\left(\frac{2}{9}\right)$ $E(X) = \frac{4}{3}$ To find the variance, we use the formula,
$Var(X) = \Sigma x^{2}p - [E(X)]^{2}$ Let’s find $\Sigma x^{2}p$ ,
$\Sigma x^{2}p = \left(-1\right)^{2}\left(\frac{1}{9}\right) + 0^{2}\left(\frac{2}{9}\right) + 1^{2}\left(\frac{1}{9}\right) + 2^{2}\left(\frac{3}{9}\right) + 3^{2}\left(\frac{2}{9}\right)$ $\Sigma x^{2}p = \frac{32}{9}$ Substitute back into the formula for variance,
$Var(X) = \Sigma x^{2}p - [E(X)]^{2}$ $Var(X) = \frac{32}{9} - \left(\frac{4}{3}\right)^{2}$ $Var(X) = \frac{16}{9}$ Therefore, the final answer is,
$Var(X) = \frac{16}{9}$