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Mean, variance and skewness of continuous variables questions
The discrete formulae with the sums replaced by integrals, plus the three averages that separate when a distribution is lopsided.
7 original questions · 28 marks · the mean, variance and skewness of continuous variables notes · Further Statistics 2
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f(x) = x/8 for 0 ≤ x ≤ 4. Find E(X) and Var(X).
Worked answer
E(X) = ∫ x²/8 dx = [x³/24] from 0 to 4 = 64/24 = 8/3, about 2.667. E(X²) = ∫ x³/8 dx = [x⁴/32] = 256/32 = 8. So Var(X) = 8 − (8/3)² = 8 − 64/9 = 8/9, about 0.889. M1 for the integrals, A1 for 8/3, A1 for 8/9.The continuous random variable X has E(X) = 8/3 and Var(X) = 8/9. Find E(3X + 2) and Var(3X + 2).
Worked answer
E(3X + 2) = 3E(X) + 2 = 3 × 8/3 + 2 = 10. Var(3X + 2) = 3²Var(X) = 9 × 8/9 = 8. B1 for 10, M1 for using a²Var(X), A1 for 8. The shift changes the mean but not the spread, and the scale factor squares in the variance.For f(x) = x/8 on 0 ≤ x ≤ 4, find the median and the mode, and describe the skew with a justification.
Worked answer
F(x) = x²/16, so the median solves m²/16 = 0.5, that is m² = 8 and m = 2√2 = 2.828. The density increases throughout the range, so the mode sits at the right-hand end, 4.
The mean is 8/3 = 2.667. Since mean 2.667 < median 2.828 < mode 4, the distribution is negatively skewed, with its tail running to the left. M1 for solving F(m) = 0.5, A1 for 2.828, B1 for the mode 4, B1 for the skew with its justification. Quote the three values in order; the skew verdict rests on that comparison.f(x) = 3(1 − x)² for 0 ≤ x ≤ 1. Find the mean, the variance and the median, and name the skew.
Worked answer
E(X) = 3∫(x − 2x² + x³) dx = 3(1/2 − 2/3 + 1/4) = 3/12 = 0.25.
E(X²) = 3∫(x² − 2x³ + x⁴) dx = 3(1/3 − 1/2 + 1/5) = 3/30 = 0.1, so Var(X) = 0.1 − 0.25² = 0.1 − 0.0625 = 0.0375.
Integrating the density, F(x) = 1 − (1 − x)³. The median solves (1 − m)³ = 0.5, so 1 − m = 0.7937 and m = 0.206.
The density falls throughout, so the mode is at 0. Then mode 0 < median 0.206 < mean 0.25, so the distribution is positively skewed. M1 for the mean integral, A1 for 0.25, M1 for E(X²) with the variance formula, A1 for 0.0375, A1 for the median 0.206, B1 for the skew. Expanding the bracket before integrating is safer than trying to integrate 3(1 − x)² x directly.For f(x) = x/8 on 0 ≤ x ≤ 4, find E(1/X).
Worked answer
E(1/X) = ∫ (1/x)(x/8) dx from 0 to 4 = ∫ 1/8 dx = [x/8] = 0.5. M1 for the integral of (1/x)f(x), A1 for the simplified integrand, A1 for 0.5. This is not 1/E(X), which would be 3/8 = 0.375. The expectation of a function is not the function of the expectation.Prove from the integral definitions that Var(aX + b) = a²Var(X) for a continuous random variable.
Worked answer
E(aX + b) = ∫(ax + b)f(x) dx = a∫xf(x) dx + b∫f(x) dx = aE(X) + b, using the fact that the density integrates to 1. Then Var(aX + b) = E([aX + b − aE(X) − b]²) = E(a²[X − E(X)]²) = a²E([X − E(X)]²) = a²Var(X). M1 for E(aX + b) = aE(X) + b, M1 for substituting into the variance definition, A1 for a²Var(X). The constant b disappears because it shifts the variable and its mean equally, leaving the deviations untouched.The continuous random variable X has f(x) = x/8 for 0 ≤ x ≤ 4, with mean 8/3 and variance 8/9. Find the three quartiles, and hence find the quartile coefficient of skewness (Q₃ − 2Q₂ + Q₁)/(Q₃ − Q₁). Compare it with the value of 3(mean − median)/σ and comment.
Worked answer
F(x) = x²/16 on the range, so each quartile solves x²/16 = p.
Q₁: x² = 4, so Q₁ = 2.
Q₂: x² = 8, so Q₂ = 2√2 = 2.828.
Q₃: x² = 12, so Q₃ = 2√3 = 3.464.
Then Q₃ − 2Q₂ + Q₁ = 3.464 − 5.657 + 2 = −0.193, and Q₃ − Q₁ = 1.464. The quartile coefficient is −0.193/1.464 = −0.132 to three decimal places.
For the other measure, σ = √(8/9) = 0.943, so 3(mean − median)/σ = 3(2.667 − 2.828)/0.943 = −0.515.
M1 for solving F(x) = p, A1 for all three quartiles, M1 for the quartile coefficient, A1 for −0.132, B1 for −0.515, B1 for the comparison.
The two disagree in size but agree in sign, and the sign is what carries the meaning: both report negative skew, matching mean < median < mode found earlier. They differ because the quartile measure reads only the middle half of the distribution and is scaled by the interquartile range, while the other uses the whole distribution through σ. Comparing the numerical values of two different coefficients of skewness is the error to avoid; only the sign transfers between them.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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