Year 12 Probability: Expected value and statistics

Once a distribution is known, you can extract summary information from it. The expected value gives a long-run average, and other statistics describe how the variable is spread around that centre.

Expected value is not always a value the variable can actually take. It is an average outcome across repeated trials.

Main formulas

$E (X) = Σ x P (X = x)$ for a discrete variable.
For continuous distributions, use integrals instead of sums.
Keep the interpretation attached to the random variable.

Worked example

Problem: A random variable

X

has distribution

P (X = 0) = 0.1

P (X = 1) = 0.3

P (X = 2) = 0.4

and

P (X = 3) = 0.2

. Find the expected value and the variance.

Calculate the expected value: $E (X) = 0 (0.1) + 1 (0.3) + 2 (0.4) + 3 (0.2) = 1.7$ .
Calculate $E (X^2)$ : $0^2 (0.1) + 1^2 (0.3) + 2^2 (0.4) + 3^2 (0.2) = 3.7$ .
Use $Var (X) = E (X^2) - (E (X))^2$ .
So $Var (X) = 3.7 - (1.7)^2 = 3.7 - 2.89 = 0.81$ .

Answer:

E (X) = 1.7

and

Var (X) = 0.81

What to watch

Use the correct probabilities with the correct values.
Keep units in mind when interpreting the result.
Do not confuse "most likely value" with "expected value".

Worked example 2

Problem: A game costs 2 dollars to play. You win 0 dollars with probability 0.5, 3 dollars with probability 0.3, and 10 dollars with probability 0.2. Find the expected net gain.

First find the expected payout: $E (Payout) = 0 (0.5) + 3 (0.3) + 10 (0.2)$ .
This gives $0 + 0.9 + 2 = 2.9$ dollars.
Now subtract the cost to play: $2.9 - 2 = 0.9$ .

Answer: the expected net gain is

0.9

dollars per game.

Interpretation matters

Expected value describes a long-run average. It is not promising that one single play will return that amount. In exam questions, the interpretation is often just as important as the arithmetic, especially in game or business contexts.

Revision focus

Probability work becomes far more reliable when you define the object of the question before calculating. Is the problem about ordered outcomes, unordered selections, a conditional event, a discrete random variable, or a continuous density? That decision controls the method. In many exam questions, the major error is not arithmetic. It is using a correct formula for the wrong kind of probability situation.

A second useful habit is to interpret the result immediately after it is found. Decide whether the answer is a count, a probability, an average outcome, or a spread measure. In continuous settings, remember that probabilities come from areas over intervals, not from single points. In discrete settings, remember that the total probability must still add to one. Those interpretation checks are often the fastest way to catch an error before it reaches the final line.