Maths › Further Statistics 2 › Mean, variance and skewness of continuous variables
Mean, variance and skewness of continuous variables
The discrete formulae with the sums replaced by integrals, plus the three averages that separate when a distribution is lopsided.
Builds on Density and distribution functions and Discrete random variables and expectation.
IN THIS TOPIC
- Find the mean, variance and E(g(X)) for a continuous variable by integration.
- Locate the mode, median and percentiles and distinguish them.
- Describe skewness from the ordering of the three averages and justify it.
WHAT YOU PROBABLY THINK
The mode of a continuous distribution is the value that occurs most often.
Sums become integrals
Every discrete formula carries over with Σ replaced by ∫ and P(X = x) replaced by f(x) dx:
The same substitution gives E(g(X)) as the integral of g(x)f(x), so there is no need to find the distribution of g(X) first. The limits are the ends of the range where f is non-zero, and the coding results, E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X), are unchanged.
WORKED EXAMPLE
Mean and variance by integration
For f(x) = 3x²/8 on 0 ≤ x ≤ 2, find the mean and variance.
E(X) = ∫ 3x³/8 dx = [3x⁴/32] from 0 to 2 = 48/32 = 1.5.
E(X²) = ∫ 3x⁴/8 dx = [3x⁵/40] from 0 to 2 = 96/40 = 2.4.
Var(X) = 2.4 − 1.5² = 0.15, so the standard deviation is about 0.387.
Three averages, and the gap between them
The mode is the value where the density is greatest, found by differentiating f or by inspecting the ends of the range. It is not a value that occurs often, since no individual value occurs at all: the opening claim borrows the discrete definition where it does not apply. The median solves F(m) = 0.5, and the mean is the integral above.
For a symmetric distribution all three coincide. When they separate, the order names the skew: mean above median above mode is positive skew, with the long tail to the right; the reverse order is negative skew. Quoting the order and naming the tail is what an examiner wants for the justification mark.
YOUR TURN
Naming the skew
For f(x) = 2(1 − x) on 0 ≤ x ≤ 1, find the mean, median and mode, and describe the skew.
Show the working
E(X) = ∫ 2x(1 − x) dx = 2(1/2 − 1/3) = 1/3.
F(x) = 2x − x², so the median solves x² − 2x + 0.5 = 0, giving m = 1 − √0.5 = 0.293.
The density falls throughout the range, so the mode is at the left-hand end, 0.
Mode 0 < median 0.293 < mean 0.333, so the distribution is positively skewed, with its tail to the right.
THE EXAM BIT
- State the integral with its limits before evaluating; the limits are the ends of the range, not zero to infinity.
- Use E(X²) − [E(X)]², and keep the square of the mean until the last line.
- For the mode, differentiate f, but check the endpoints too: a monotonic density peaks at an end.
- Justify skewness by the order of the three averages, and name which side the tail lies on.
CHECK YOURSELF
For f(x) = 1/4 on 0 ≤ x ≤ 4, find E(X) and E(X²).
Show a hint
Integrate x and x² against the constant density.
Show the answer
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Replace sums with integrals: E(X) is the integral of xf(x), E(g(X)) the integral of g(x)f(x), and Var(X) = E(X²) − [E(X)]².
Mode is where f peaks, median solves F(m) = 0.5, and the order of mode, median and mean names the skew and points at the tail.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Find the mean, variance and E(g(X)) for a continuous variable by integration.
- Locate the mode, median and percentiles and distinguish them.
- Describe skewness from the ordering of the three averages and justify it.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.