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.
- Apply the coding results E(aX + b) and Var(aX + b) in the continuous case.
COMMON MISCONCEPTION
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 booklet gives both under Continuous distributions, with the variance written as ∫x²f(x) dx − μ², and it gives E(g(X)) = ∫g(x)f(x) dx as well. The same substitution gives E(g(X)) as the integral of g(x)f(x), so you never need the distribution of g(X) first. Limits are the ends of the range where f is non-zero. The coding results survive untouched, so E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X) exactly as before.
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.
Mode, median and the percentiles between them
The mode is the value where the density is greatest. Differentiate f and solve f'(x) = 0, but check the ends of the range as well, because a density that falls throughout its range peaks at its left-hand endpoint and the calculus will not find it. A mode is not a value that occurs often, since no individual value occurs at all; that is a discrete definition borrowed into a place it does not fit.
The median solves F(m) = 0.5. Any percentile works the same way, with the pth percentile solving F(x) = p/100, so the quartiles come from 0.25 and 0.75. When F is piecewise, check which branch the answer belongs in before you solve, and reject a root that lands outside the range.
WORKED EXAMPLE
A percentile from a piecewise F
F(x) = x²/16 on 0 ≤ x ≤ 4, with the usual flat branches. Find the interquartile range.
Lower quartile: q²/16 = 0.25 gives q² = 4, so q1 = 2.
Upper quartile: q²/16 = 0.75 gives q² = 12, so q3 = 3.46 to three significant figures.
The interquartile range is 1.46. Both roots lie in 0 to 4, and the negative roots are discarded.
Three averages, and the gap between them
For a symmetric distribution the mean, median and mode coincide. When they separate, their order names the skew. Mean above median above mode is positive skew, with the long tail running to the right. The reverse order is negative skew. Quoting the order and naming the tail is what earns the justification mark.
Take that ordering as a working sign rather than a definition. Skew is about which tail is the longer, and the rule reads correctly for the single-peaked densities this course sets while it can mislead for stranger ones. The reverse reading is weaker still. A symmetric density does put all three averages in the same place, but all three landing together is not a proof that a density is symmetric.
GUIDED PRACTICE
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 sits 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.
ASSESSMENT FOCUS
- 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 intact until the last line.
- For the mode, differentiate f, then 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
E(X) = ∫ x/4 dx = [x²/8] from 0 to 4 = 2. E(X²) = [x³/12] from 0 to 4 = 64/12 = 16/3, so Var(X) = 16/3 − 4 = 4/3.
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.
The mode is where f peaks, endpoints included, any percentile solves F(x) = p/100, and the order of the three averages names the skew.
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.
Or read them with their worked answers on the mean, variance and skewness of continuous variables questions page.
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.
- Apply the coding results E(aX + b) and Var(aX + b) in the continuous case.
Open the full revision checklist to see every objective in the course in one place.