## Random Variables

A discrete random variable can only take particular values. They are numbers that are countable and are described by a probability distribution. They can be represented by a table, a graph or a formula which takes into account of the probability p(x) for every value of x.

### Probability Functions

Now let’s look at probability functions. There are 3 rules that we need to remember when we are doing a probability functions question.

• The value of p(x) must be greater than 0 and less than 1

$0 \leq p(x) \leq 1$ for all of x

• The sum of all values of p(x) must be exactly 1

$\sum\limits_{x}{p(x)}=1$

• In order to determine the probability of x in the interval from a to b, the values of p(x) from x=a to x=b are added

$a \leq pr(x) \leq b =\sum\limits_{x=a}^{x=b}{p(x)}$

## Measure of Centre

The three main measures of centre are mean, median and mode.

### Mean

The mean is also known as the expected value is used to calculate the average value we would expect to occur from the function in the long term and is calculated with the formula:

$E(x)=\sum\limits_{x}^{x} \times p(x)$

### Median

Median is the middle value of the distribution in which 50% of the distribution is larger than it and 50% is less than it.

### Mode

The mode is the discrete value which is most commonly occurring discrete value.

### Practice Exam Question

Question 7 from the 2013 VCAA exam 1

As the sum of all values of p(x) must equal to 1, in order to solve for p, we need to add all the values of Pr(X=x) together and make them equal to one.

$\sum{Pr(X=x)}=0.2+0.6p^{2}+0.1+1-p+0.1=1 \newline 0.6p^{2}-p+0.4=0 \newline p= 1, \frac{2}{3}$

#### Calculating Measure of Centre

This question is related to the previous question that we have gone through. First, we need to substitute this p value into the table.

 0 1 2 3 4 Pr( 0.2 0.267 0.1 0.333 0.1

In order to find the expected value. We need to find the sum of all the values of x times by the probability that it takes, as the formula for the mean above shows us.

$E(x)=\sum\limits_{x}^{x} \times p(x) \newline = 0 \times 0.2 +1 \times 0.267 +2 \times 0.1 +3 \times 0.333 +4 \times 0.1\newline =1.86$

The mode for this set is 3 as that is the value most likely to occur.

Looking again at the original table we can see that 46.7% of the time the value of the number will be less than two, but the probability that is less or equal to two is 56.7%. This means that 2 sits on the range of 50% of the distribution is larger than it and 50% is smaller. Therefore it is the median value.