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,

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,

Therefore, the final answer is,

(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,

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,

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,

Therefore, the final answer is,

(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,
$$Var(X) = \frac{16}{9}$$