Questions on mean, variance and skewness of continuous variables

Question 1

f(x) = x/8 for 0 ≤ x ≤ 4. Find E(X) and Var(X).

Accepted 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.

Question 2

The continuous random variable X has E(X) = 8/3 and Var(X) = 8/9. Find E(3X + 2) and Var(3X + 2).

Accepted 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.

Question 3

State how the mode of a continuous random variable is found from its density, and how the median is found from its distribution function.

Accepted Answer

The mode is the value of x at which the density f is greatest across the range: either a stationary value found by solving f′(x) = 0, or an endpoint of the range where f is increasing or decreasing throughout. B1. The median m is the solution of F(m) = 0.5. B1.
Solving f(m) = 0.5 finds a point where the density takes the value one half and says nothing about where the area splits.

Question 4

The continuous random variable X has density f(x) = 1/4 for 2 ≤ x ≤ 6 and zero elsewhere. Find E(X) and E(X²) by integration.

Accepted Answer

E(X) = ∫x/4 dx from 2 to 6 = [x²/8] = 36/8 − 4/8 = 32/8 = 4. M1 for the integral of xf(x), A1 for 4.
E(X²) = ∫x²/4 dx from 2 to 6 = [x³/12] = 216/12 − 8/12 = 208/12 = 52/3, about 17.33. A1.
Together they give Var(X) = 52/3 − 16 = 4/3, which agrees with the continuous uniform result (6 − 2)²/12 = 4/3, so the two integrals check each other.

Question 5

For f(x) = x/8 on 0 ≤ x ≤ 4, find the median and the mode, and describe the skew with a justification.

Accepted 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.

Question 6

f(x) = 3(1 − x)² for 0 ≤ x ≤ 1. Find the mean, the variance and the median, and name the skew.

Accepted 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.

Question 7

For f(x) = x/8 on 0 ≤ x ≤ 4, find E(1/X).

Accepted 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.

Question 8

Prove from the integral definitions that Var(aX + b) = a²Var(X) for a continuous random variable.

Accepted 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).

Question 9

The continuous random variable X has density f(x) = 3x²/64 for 0 ≤ x ≤ 4 and zero elsewhere. Find E(X) and Var(X).

Accepted Answer

E(X) = ∫x × 3x²/64 dx from 0 to 4 = (3/64)[x⁴/4] = (3/64)(64) = 3. M1 for the integral of xf(x), A1 for 3.
E(X²) = ∫x² × 3x²/64 dx = (3/64)[x⁵/5] = (3/64)(1024/5) = 9.6. M1 for the integral of x²f(x).
Var(X) = 9.6 − 3² = 0.6. A1.
Check the density first: ∫3x²/64 dx from 0 to 4 = [x³/64] = 1, so it is a valid density.

Question 10

The waiting time X minutes between arrivals is modelled by an exponential distribution with density f(x) = λe^−λx for x ≥ 0 and distribution function F(x) = 1 − e^−λx. Show by integration that E(X) = 1/λ, and find the median of X in terms of λ.

Accepted Answer

E(X) = ∫λxe^−λx dx from 0 to ∞. Integrate by parts with u = x and dv = λe^−λx dx, so v = −e^−λx:
E(X) = [−xe^−λx] from 0 to ∞ + ∫e^−λx dx from 0 to ∞. M1 for the parts split, A1 for the two pieces.
The bracket is 0 at both ends, since xe^−λx → 0 as x → ∞, and the remaining integral is [−e^−λx/λ] from 0 to ∞ = 1/λ. So E(X) = 1/λ. A1.
The median solves F(m) = 0.5, that is 1 − e^−λm = 0.5, so e^−λm = 0.5 and λm = ln 2, giving m = (ln 2)/λ. M1 for F(m) = 0.5, A1 for (ln 2)/λ.
Since ln 2 = 0.693 < 1, the median lies below the mean at every rate, which is the signature of positive skew: the long right tail pulls the mean above the point that splits the area in half.

Question 11

The continuous random variable X has distribution function F(x) = (x − 2)³/27 for 2 ≤ x ≤ 5, with F(x) = 0 below 2 and F(x) = 1 above 5. Find the 90th percentile of X and the mode of X.

Accepted Answer

The 90th percentile solves F(x) = 0.9, so (x − 2)³ = 27 × 0.9 = 24.3. M1 for setting F equal to 0.9.
Taking the cube root, x − 2 = 2.8965, so the percentile is 4.90 to two decimal places. A1. It lies inside the range 2 to 5, as a percentile must.
For the mode, differentiate: f(x) = 3(x − 2)²/27 = (x − 2)²/9 on the range. M1.
That is increasing throughout 2 ≤ x ≤ 5, rising from 0 to 1, so the density is greatest at the right-hand end and the mode is 5. A1.
Solving f′(x) = 0 here gives x = 2, which is where the density is smallest, not largest: with a density that is monotonic across the range, the mode sits at an endpoint and the stationary point is no guide.

Question 12

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.

Accepted 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.

Question 13

The continuous random variable X has density f(x) = kx(4 − x) for 0 ≤ x ≤ 4 and zero elsewhere. Find k, E(X) and Var(X), and state the skew of the distribution with a reason.

Accepted Answer

∫(4x − x²) dx from 0 to 4 = [2x² − x³/3] = 32 − 64/3 = 32/3, so k × 32/3 = 1 and k = 3/32. M1 for setting the total area to 1, A1 for k.
E(X) = (3/32)∫(4x² − x³) dx = (3/32)[4x³/3 − x⁴/4] = (3/32)(256/3 − 64) = (3/32)(64/3) = 2. M1 for the integral of xf(x), A1 for 2.
E(X²) = (3/32)∫(4x³ − x⁴) dx = (3/32)[x⁴ − x⁵/5] = (3/32)(256 − 204.8) = (3/32)(51.2) = 4.8, so Var(X) = 4.8 − 2² = 0.8. A1.
The density is symmetric about x = 2, since replacing x by 4 − x leaves x(4 − x) unchanged. So the mean, the median and the mode all sit at 2 and the skew is zero. B1 for zero skew with the symmetry as the reason.

Question 14

An exponential distribution has density f(x) = λe^−λx for x ≥ 0, mean 1/λ and median (ln 2)/λ. Show by integration that its variance is 1/λ², compare the mean with the median and name the skew. The time in days between breakdowns of a machine is modelled by such a distribution with a mean of 40 days. Find the probability that the machine runs for more than 60 days without a breakdown.

Accepted Answer

E(X²) = ∫λx²e^−λx dx from 0 to ∞. Integrating by parts with u = x² and dv = λe^−λx dx gives [−x²e^−λx] + ∫2xe^−λx dx; the bracket vanishes at both ends and the integral that remains is (2/λ)∫λxe^−λx dx = (2/λ)E(X) = 2/λ². M1 for parts, A1 for 2/λ².
So Var(X) = 2/λ² − (1/λ)² = 1/λ²: the standard deviation equals the mean. A1.
Since ln 2 = 0.693, the median (ln 2)/λ is about seven tenths of the mean 1/λ, so the median lies below the mean and the distribution is positively skewed, with a long tail of rare very long waits. B1 for the comparison with the skew named.
For the machine, the mean is 1/λ = 40, so λ = 1/40 per day and P(X > 60) = e^−60/40 = e^−1.5 = 0.223. M1 for the tail, A1 for 0.223.
Note that the mean of 40 days does not make 40 the typical wait: the median is only 0.693 × 40 = 27.7 days, so more than half of the gaps between breakdowns are shorter than four weeks.

Question 15

A continuous random variable X has mean μ and variance σ². Show that E(X²) = σ² + μ². Given that a variable has mean 5 and variance 4, find E((X − 3)²), and find the value of a for which E((X − a)²) is least.

Accepted Answer

By definition Var(X) = E(X²) − [E(X)]², so σ² = E(X²) − μ² and E(X²) = σ² + μ². B1 for the variance formula rearranged to the result.
Expanding, E((X − a)²) = E(X² − 2aX + a²) = E(X²) − 2aE(X) + a² = σ² + μ² − 2aμ + a² = σ² + (μ − a)². M1 for expanding the expectation, A1 for σ² + (μ − a)².
With μ = 5 and σ² = 4, taking a = 3 gives 4 + (5 − 3)² = 8. A1.
The expression σ² + (μ − a)² is least when (μ − a)² = 0, that is a = μ = 5, where it takes the value σ² = 4. A1.