Questions on continuous random variables: density and distribution functions

Question 1

Explain why P(X = 1.5) = 0 for a continuous random variable, and what follows for strict and non-strict inequalities.

Accepted Answer

Probability is the area under the density, and the area over a single point is zero, so no individual value carries any probability. It follows that P(X < a) and P(X ≤ a) are equal, so the inequality signs make no difference. B1 for the zero area over a point, B1 for the two inequalities agreeing.

Question 2

f(x) = k(4 − x²) for 0 ≤ x ≤ 2 and zero elsewhere. Find k.

Accepted Answer

The integral of 4 − x² from 0 to 2 is [4x − x³/3] = 8 − 8/3 = 16/3. Setting k × 16/3 = 1 gives k = 3/16. M1 for integrating and setting the total area to 1, A1 for 16/3, A1 for k = 3/16. The density is non-negative throughout the interval, so it is valid.

Question 3

State how the distribution function F of a continuous random variable is obtained from its density f, and how f is obtained from F.

Accepted Answer

F is the integral of f from the bottom of the range up to x: F(x) = ∫f(t) dt taken from −∞ to x, which for a density that starts at a means ∫ from a to x. B1. Going the other way, f is the derivative of F: f(x) = dF/dx. B1. Where a question gives F, use it: probabilities over an interval are then a subtraction rather than an integral.

Question 4

The waiting time X hours between arrivals at a hut is modelled by the distribution function F(x) = 1 − e^−0.4x for x ≥ 0, and F(x) = 0 for x < 0. Find the density of X and find P(X > 5).

Accepted Answer

Differentiating F on x ≥ 0 gives f(x) = 0.4e^−0.4x, and f(x) = 0 for x < 0. M1 for differentiating, A1 for the density with its range.
P(X > 5) = 1 − F(5) = e⁻² = 0.135. A1.
This is the exponential distribution with rate λ = 0.4 arrivals per hour, so the mean wait is 1/0.4 = 2.5 hours.

Question 5

The random variable X has density f(x) = (3/16)(4 − x²) for 0 ≤ x ≤ 2 and zero elsewhere. Find F(x) and hence P(X < 1).

Accepted Answer

Integrate the density from the bottom of the range: F(x) = (3/16)(4x − x³/3) = 3x/4 − x³/16 for 0 ≤ x ≤ 2, with F = 0 below 0 and F = 1 above 2. A distribution function must be defined everywhere, so both of those branches are required.
Check F(2) = 1.5 − 0.5 = 1, which every F must satisfy at the top of its range.
Then P(X < 1) = F(1) = 0.75 − 0.0625 = 0.6875. M1 for integrating the density, A1 for 3x/4 − x³/16, B1 for the branches outside the range, A1 for 0.6875.

Question 6

The random variable X has density f(x) = x/8 for 0 ≤ x ≤ 4 and zero elsewhere. Find F(x), the median and the lower quartile.

Accepted Answer

Integrating, F(x) = x²/16 on the interval. Quartiles come from solving F = the required proportion, never from the density. The median solves m²/16 = 0.5, so m² = 8 and m = 2√2 = 2.828. The lower quartile solves q²/16 = 0.25, so q² = 4 and q = 2. M1 A1 for F(x) = x²/16, A1 for the median 2.828, A1 for the lower quartile 2. Both lie inside 0 to 4, and the lower quartile lies below the median. The negative roots are discarded because the density is zero below 0.

Question 7

A variable has F(x) = x²/16 on 0 ≤ x ≤ 4. Find its density and P(1 < X < 3).

Accepted Answer

Differentiating, f(x) = x/8 on the interval and zero elsewhere. P(1 < X < 3) = F(3) − F(1) = 9/16 − 1/16 = 8/16 = 0.5. B1 for f(x) = x/8, M1 for F(3) − F(1), A1 for 0.5. Subtracting values of F is quicker than integrating.

Question 8

The random variable X has density f(x) = x/8 for 0 ≤ x ≤ 4, so that F(x) = x²/16 there. Find the distribution function and the density of Y = X², and describe the distribution of Y.

Accepted Answer

Work on F, not on f. X runs from 0 to 4, so Y = X² runs from 0 to 16. B1 for the range of Y carried through the same rule. Squaring is increasing on the range, so the event X² ≤ y says the same thing as X ≤ √y, and for 0 ≤ y ≤ 16, F_Y(y) = P(X ≤ √y) = (√y)²/16 = y/16, with 0 below the range and 1 above it. M1 for rewriting the event about X, A1 for F_Y in full. Differentiating, f_Y(y) = 1/16 for 0 ≤ y ≤ 16: the square of this X is uniform on the interval. A1 for the flat density named as uniform. Two checks: a constant 1/16 across a width of 16 has area 1, and E(Y) = E(X²) = ∫ x² × x/8 dx from 0 to 4 = 8, the midpoint of 0 to 16, where a uniform mean belongs. Substituting √y into the density instead misses the chain-rule factor and is exactly the error working through F avoids.

Question 9

The random variable X has density f(x) = k(x − 1)(3 − x) for 1 ≤ x ≤ 3 and zero elsewhere. Show that k = 3/4, and find P(X > 2.5).

Accepted Answer

Expand first: (x − 1)(3 − x) = −x² + 4x − 3. Integrating from 1 to 3 gives [−x³/3 + 2x² − 3x], which is (−9 + 18 − 9) − (−1/3 + 2 − 3) = 0 − (−4/3) = 4/3. M1 for integrating, A1 for 4/3.
Total probability is 1, so k × 4/3 = 1 and k = 3/4. A1.
For the tail, integrate over 2.5 to 3: (3/4)[0 − (−0.2083…)] = (3/4)(0.2083…) = 5/32, that is 0.15625. M1 for the tail integral, A1 for 5/32.
Check the density is a legitimate one: (x − 1)(3 − x) is non-negative for every x from 1 to 3, so f ≥ 0 across the range, and it now integrates to 1. Both conditions are needed. Integrating from 0 rather than from 1 adds area where the density is zero and produces the wrong k.

Question 10

The continuous random variable X has distribution function F(x) = 0 for x < 0, F(x) = x³/27 for 0 ≤ x ≤ 3, and F(x) = 1 for x > 3. Find the density of X, the median of X, and P(1 < X < 2).

Accepted Answer

Differentiating on the range, f(x) = x²/9 for 0 ≤ x ≤ 3 and zero elsewhere. B1. The median solves F(m) = 0.5, that is m³/27 = 0.5, so m³ = 13.5 and m = 2.381 to three decimal places. M1 for setting F(m) = 0.5, A1 for 2.381. It lies between 0 and 3, as it must, and above the midpoint 1.5 because the density rises across the range. P(1 < X < 2) = F(2) − F(1) = 8/27 − 1/27 = 7/27, or 0.259. A1. Solving f(m) = 0.5 instead answers nothing: the median is a property of the accumulated area, so it comes from F.

Question 11

Calls reach a helpline at random, singly and independently, at a mean rate of 5 per hour. The wait T hours until the next call is modelled by an exponential distribution. Write down the value of λ, and find the probability that the wait is less than 6 minutes and the probability that it is more than half an hour.

Accepted Answer

The rate is λ = 5 per hour, so P(T > t) = e^−5t with t measured in hours. B1. Convert the windows into hours before substituting: 6 minutes is 0.1 h and half an hour is 0.5 h. M1. P(T < 0.1) = 1 − e^−0.5 = 0.393. A1. P(T > 0.5) = e^−2.5 = 0.0821. A1. λ and t must be counted in the same unit. Leaving the first window as 6 gives e⁻³⁰ and a probability of essentially 1, which the mean wait of 1/5 h = 12 minutes shows at once to be wrong.

Question 12

The random variable X has density f(x) = x for 0 ≤ x < 1 and f(x) = 2 − x for 1 ≤ x ≤ 2, and zero elsewhere. Verify that f is a valid density, find F(x) in full, and find the upper quartile of X, giving an exact answer.

Accepted Answer

Validity. f is non-negative on the whole range, and the two triangles each have area 1/2, so the total area is 1. Both conditions are needed; area alone is not enough. The distribution function. For 0 ≤ x < 1, F(x) = x²/2. For the second branch, start from the 1/2 already accumulated rather than integrating from 0: F(x) = 1/2 + [2t − t²/2] from 1 to x = 2x − x²/2 − 1. Two checks. At x = 1 both branches give 0.5, so F is continuous, and F(2) = 4 − 2 − 1 = 1 as it must be. The upper quartile. Since F(1) = 0.5 < 0.75, the quartile lies in the upper branch. Solve 2q − q²/2 − 1 = 0.75, that is q² − 4q + 3.5 = 0, giving q = 2 ± √2/2. The root 2 + √2/2 ≈ 2.71 lies outside the range, so the upper quartile is q = 2 − √2/2, about 1.293. B1 for the validity check, M1 for integrating the first branch, A1 for both branches of F, B1 for choosing the upper branch, dM1 for solving the quadratic, A1 for q = 2 − √2/2. Say why the other root is rejected; a bare pair of roots does not finish the question.

Question 13

A driver arriving at a car park is waved straight in with probability 0.3, waiting T = 0 minutes; otherwise the wait is spread evenly, with density f(t) = k for 0 < t ≤ 14. Explain why k must satisfy 0.3 + 14k = 1, and hence find k, P(T ≤ 6) and the median wait.

Accepted Answer

T is part discrete and part continuous: it carries a lump of probability 0.3 at the single value 0, and a density over the positive waits. The lump and the area under the density share one total of 1, so 0.3 + ∫k dt over 0 to 14 = 0.3 + 14k = 1. B1 for the total-probability argument across both parts. So k = 0.7/14 = 0.05. A1. The distribution function steps to 0.3 at t = 0 and then climbs: F(t) = 0.3 + 0.05t on 0 ≤ t ≤ 14. M1 for building F with the step included. P(T ≤ 6) = 0.3 + 0.05 × 6 = 0.6. B1. The median solves 0.3 + 0.05m = 0.5, so m = 4 minutes, landing on the density rather than on the lump because F has already passed 0.3 at zero but not yet 0.5. A1. Forgetting the lump and solving 0.05m = 0.5 gives 10, a median for a distribution this variable does not have.

Question 14

The lifetime T hours of an electronic component is modelled by an exponential distribution with mean 800 hours. Find the probability that a new component lasts beyond 1000 hours. Show that P(T > 500 + t | T > 500) = P(T > t) for every t > 0, and state what this says about a component that has already run for 500 hours. Explain why the model would be unsuitable for a part that wears.

Accepted Answer

The mean of an exponential distribution is 1/λ, so λ = 1/800 per hour and P(T > t) = e^−t/800. B1.
P(T > 1000) = e^−1.25 = 0.287. M1 for the tail, A1 for 0.287.
For the conditional probability, T > 500 + t already implies T > 500, so the intersection is the smaller event and
P(T > 500 + t | T > 500) = P(T > 500 + t)/P(T > 500) = e^{−(500 + t)/800} ÷ e^−500/800. M1 for the conditional probability as a quotient of the two tails.
Subtracting the indices leaves e^−t/800, which is P(T > t). A1.
So a component that has already run for 500 hours faces exactly the distribution of remaining life it faced when new: the model remembers nothing about its age. That is why it does not suit a part that wears, where the chance of failing in the next month rises as the part ages. B1 for the interpretation with the unsuitability explained.
Note that P(T > 1500 | T > 500) is 0.287 again, the same figure as for a new component over 1000 hours, so the second calculation is one already done.

Question 15

The random variable X has density f(x) = 3x² for 0 ≤ x ≤ 1 and zero elsewhere. Find the distribution function of X, and hence find the distribution function and the density of Y = 4X + 1, stating the range of Y in each case.

Accepted Answer

Integrating from the bottom of the range, F_X(x) = x³ for 0 ≤ x ≤ 1, with 0 below and 1 above. M1 for integrating, A1 for x³.
X runs from 0 to 1 and Y = 4X + 1 is increasing, so Y runs from 1 to 5. B1 for the range carried through the same rule.
Work on F rather than on f. For 1 ≤ y ≤ 5 the event Y ≤ y says the same thing as X ≤ (y − 1)/4, so
F_Y(y) = F_X((y − 1)/4) = ((y − 1)/4)³ = (y − 1)³/64, with 0 below 1 and 1 above 5. M1 for rewriting the event about X, A1 for F_Y.
Differentiating, f_Y(y) = 3(y − 1)²/64 for 1 ≤ y ≤ 5. A1.
Two checks: F_Y(5) = 64/64 = 1, and the density integrates back to F_Y. Substituting (y − 1)/4 straight into f without the factor from differentiating the inside gives 3(y − 1)²/16, which is four times too large and integrates to 4 rather than 1.