1The idea in one paragraph

A random variable X is a number decided by chance. If its values can be listed, such as 0, 1, 2, 3 goals, it is discrete, and each value has a probability. If it can take any value in an interval, such as a waiting time, it is continuous: a single exact value has probability zero, and probability is the area under a curve called the probability density function, f(x). Either way, the same few numbers summarise it. The mean E(X) is its balance point. The variance Var(X) is the average squared distance from that mean, and its square root, the standard deviation, is a typical distance. For a continuous variable you also find the mode, where f is highest, and the median, which splits the area in half. Finally, if you convert X with a linear rule, aX + b, the mean follows the same rule but the variance only feels the a, squared.

2Discrete random variables: mean and variance

A football team's goals per match, X, have this distribution:

The expected value or mean is the probability-weighted average, E(X) = Σ x P(X = x), which you met in 4.7.

No match ends with 1.5 goals; 1.5 is the long-run average, the point where the bar chart in Figure 1 balances.

The variance measures spread: the expected squared distance from the mean, Var(X) = E[(X − μ)²], where μ = E(X). The formula booklet gives it two ways:

Var(X) = Σ (x − μ)² P(X = x) = Σ x² P(X = x) − μ², that is, Var(X) = E(X²) − [E(X)]².

The second form is almost always quicker. E(X²) means the expected value of the square of X: square each value, weight by its probability, add.

Notice that E(X²) = 3.5 is not [E(X)]² = 2.25; the gap between them is the variance. The variance is in squared units (goals²), so the standard deviation, back in goals, is the one you interpret. Figure 1 marks both.

A distribution given by a formula (Paper 1). The random variable X has P(X = x) = kx² for x ∈ {1, 2, 3}. Find k, E(X) and Var(X).

The standard deviation is √19 ÷ 7 ≈ 0.623. Leave Paper 1 answers exact unless asked otherwise.

3Fair games

If X is a player's gain in a game (winnings minus the fee), the game is fair when E(X) = 0: in the long run nobody profits. It is not about winning as often as losing.

A charity stall charges $3 to spin a wheel. The wheel pays $10 with probability 0.1, $4 with probability 0.3, and nothing otherwise. Let W be the amount paid out.

The game is not fair: it favours the stall. It would be fair at a fee of $2.20. Insurers, casinos and lotteries price the same way, with E(gain) negative for the customer. The spread of the payout is E(W²) = 100(0.1) + 16(0.3) = 14.8, so Var(W) = 14.8 − 2.2² = 9.96; section 8 shows the gain W − 3 has the same variance.

4Continuous random variables and the probability density function

A continuous variable takes too many values to list, so instead a probability density function (pdf) f(x) is a curve whose area gives probability:

P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx, with f(x) ≥ 0 for all x and ∫ f(x) dx = 1 over all x (from −∞ to ∞).

Because single points have zero probability, P(X < 1) = P(X ≤ 1) for a continuous variable (not for a discrete one). And f(x) is not a probability; it is a density and can exceed 1: a variable spread evenly over [0, 0.5] has f(x) = 2 there, so that the area is 1. Usually f is a formula on an interval and 0 elsewhere, so "from −∞ to ∞" becomes "over that interval".

Example · finding k (Paper 1). A continuous random variable X has pdf f(x) = kx²(3 − x) for 0 ≤ x ≤ 3, and f(x) = 0 otherwise. Find k, then P(X < 1).

Figure 2 shows the curve with the total area of 1 and the shaded area 1/9. Also check the other condition: on [0, 3], x² ≥ 0 and 3 − x ≥ 0, so f(x) ≥ 0. A question saying "show that f is a valid pdf" wants both checks.

Similarly, P(1 < X < 2) = (4/27)[(8 − 4) − (1 − 1/4)] = (4/27)(13/4) = 13/27. On Paper 2 you would get 0.481 by numerical integration on the GDC; on Paper 1 you do exactly the working above.

5Mode and median of a continuous random variable

The guide defines both.

• The mode is a value of x at which the pdf f(x) has its maximum value. For a smooth curve, find it by solving f′(x) = 0 and checking it is a maximum; if f is increasing or decreasing across the whole interval, the maximum is at an endpoint.
• The median m is the value that splits the area in half: ∫ f(x) dx from −∞ to m = 1/2.

Mode. For f(x) = (4/27)(3x² − x³):

Justify the maximum: f′ changes from positive to negative at 2, or f(2) exceeds f at both endpoints, which are 0.

Median. You need (4/27)[m³ − m⁴/4] = 1/2, a quartic. On Paper 1 a question would choose a pdf where this comes out nicely (section 7 shows one); here it does not, so it is a Paper 2 job.

Reject any GDC solution outside [0, 3]. Figure 3 marks the mode, the median and the mean (found in the next section) on the same curve.

6Mean, variance and standard deviation of a continuous random variable

The continuous formulas are the discrete ones with the sum replaced by an integral and P(X = x) replaced by f(x) dx:

E(X) = μ = ∫ x f(x) dx and Var(X) = ∫ (x − μ)² f(x) dx = ∫ x² f(x) dx − μ², integrating over all x. From the formula booklet.

So E(X²) = ∫ x² f(x) dx and, exactly as before, Var(X) = E(X²) − [E(X)]². For the same pdf:

Multiply f(x) by x before integrating; ∫ f(x) dx on its own is just 1.

Look again at Figure 3. The mean 1.8 is below the median 1.84, which is below the mode 2. The curve has a long tail to the left, a negatively skewed distribution, and a tail drags the mean towards it more than it drags the median, because the mean weights each value by how far out it is. For a symmetric pdf all three coincide. This ordering is a useful check on your answers.

7Piecewise probability density functions

The guide includes piecewise functions: f is given by one rule on one interval and another rule on the next. Every property still holds, but each integral must be split at the join.

A random variable X has

f(x) = x/6 for 0 ≤ x ≤ 2, f(x) = (6 − x)/12 for 2 < x ≤ 6, and f(x) = 0 otherwise.

Figure 4 shows it: a triangle, rising in teal and falling in amber.

Check it is valid. Both pieces are non-negative on their intervals, and

Mode. The highest point is the peak of the triangle, f(2) = 1/3, so the mode is 2.

Median. The first piece holds only 1/3 of the area, less than 1/2, so the median is in the second piece. You need a further 1/2 − 1/3 = 1/6 of area beyond x = 2.

Deciding which piece holds the median before integrating is worth a mark.

Mean and variance. Split each integral at 2.

So mode 2 < median 2.54 < mean 2.67: this time the long tail is on the right, a positively skewed distribution, and the order reverses.

8Linear transformations: E(aX + b) and Var(aX + b)

If every value of X is multiplied by a and then has b added, the new variable is aX + b.

E(aX + b) = aE(X) + b and Var(aX + b) = a² Var(X). From the formula booklet.

Why. For a discrete X, E(aX + b) = Σ (ax + b)P(X = x) = aΣ x P(X = x) + bΣ P(X = x) = aE(X) + b, because the probabilities add to 1. The same step works with integrals. For the variance, each value's distance from the mean becomes (ax + b) − (aμ + b) = a(x − μ): the b cancels, and the distance is multiplied by a. Squaring gives a², so Var(aX + b) = a²Var(X), and the standard deviation becomes |a|σ.

Figure 5 shows it with the goals from section 2. A fantasy league scores a team 2 points per goal plus 1 point for playing, so its points are Y = 2X + 1.

Changing units. July maximum temperatures X °C have mean 18 and standard deviation 4. In °F, F = 1.8X + 32, so E(F) = 1.8(18) + 32 = 64.4 °F, Var(F) = 1.8² × 16 = 51.84 and the standard deviation is 7.2 °F. The + 32 moves the mean but not the spread.

Three cases that catch people. Var(X − 5) = Var(X), which is why the wheel's gain W − 3 has variance 9.96. Var(−X) = (−1)²Var(X) = Var(X). And Var(3 − 2X) = (−2)²Var(X) = 4Var(X), not −2Var(X) and not 4Var(X) + 3.