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Continuous Random Variables and Probability Density Functions — Answers

Answer sheet with worked solutions for the Year 12 statistics worksheet.

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Success TutoringWorksheetsAnswer sheetYear 12 · Mathematics · Statistics

Continuous Random Variables and Probability Density Functions — Answers

Part A: Probability density functions

A continuous random variable can take any value in an interval (a time, a mass, a length). It is described by a probability density function (pdf) f(x), with f(x) ≥ 0 everywhere and total area 1: ∫ f(x) dx = 1 over all x.

Probabilities are areas: P(a ≤ X ≤ b) = ∫ from a to b of f(x) dx. Because a single point has no area, P(X = a) = 0, and ≤ and < give the same probability.

Worked example. f(x) = kx for 0 ≤ x ≤ 2. ∫ from 0 to 2 of kx dx = 2k = 1, so k = . Then P(X < 1) = ∫ from 0 to 1 of dx = [frac 1/4].

1.X has probability density function f(x) = kx2 for 0 ≤ x ≤ 3 (and 0 elsewhere). Find k.

Answer:

  1. The total area must be 1: ∫ from 0 to 3 of kx2 dx = k[] = 9k = 1
  2. k = [frac 1/9]
2.For the same X, with f(x) = on 0 ≤ x ≤ 3, find P(X ≤ 2).

Answer:

  1. P(X ≤ 2) = ∫ from 0 to 2 of dx = evaluated from 0 to 2
  2. = [frac 8/27] (about 0.296)
3.For a continuous random variable X, P(X = 1.5) = 0.

Answer: True

The probability is the area under the density curve, and the area above a single point has zero width. So P(X ≤ 1.5) = P(X < 1.5).

4.Which function is a valid probability density function on 0 ≤ x ≤ 1 (and 0 elsewhere)?
  1. Af(x) = 2x
  2. Bf(x) = x
  3. Cf(x) = 3x2 − 1
  4. Df(x) = 1.5

Answer: f(x) = 2x

∫ from 0 to 1 of 2x dx = 1 and 2x ≥ 0, so it is valid. The area under f(x) = x is ; 3x2 − 1 is negative for small x (and its area is 0); the area under 1.5 is 1.5.

5.X has density f(x) = (4 − x) for 0 ≤ x ≤ 4. Find P(1 ≤ X ≤ 3).

Answer:

  1. P = ∫ from 1 to 3 of (4 − x) dx = [4x − ] from 1 to 3
  2. = ((12 − 4.5) − (4 − 0.5)) = (7.5 − 3.5)
  3. = = [frac 1/2]

Part B: Mean, variance and median

Mean: E(X) = μ = ∫ x f(x) dx. Variance: Var(X) = E(X2) − μ2, where E(X2) = ∫ x2 f(x) dx. Standard deviation: σ = √Var(X).

The median m splits the area in half: ∫ from the lowest value to m of f(x) dx = .

Worked example. f(x) = 2x on 0 ≤ x ≤ 1. E(X) = ∫ 2x2 dx = ; E(X2) = ∫ 2x3 dx = ; Var(X) = − = . Median: m2 = , so m = √0.5 ≈ 0.707.

6.For f(x) = on 0 ≤ x ≤ 3, find E(X).

Answer: 2.25

  1. E(X) = ∫ from 0 to 3 of x × dx = from 0 to 3
  2. = = = 2.25
7.For the same distribution, find Var(X).

Answer: 0.3375

  1. E(X2) = ∫ from 0 to 3 of dx = from 0 to 3 = = 5.4
  2. Var(X) = E(X2) − [E(X)]2 = 5.4 − 2.252 = 5.4 − 5.0625
  3. = 0.3375 (that is, )
8.For the same distribution, find the median, correct to 2 decimal places.

Answer: 2.38

  1. Solve ∫ from 0 to m of dx = : =
  2. m3 = 13.5, so m = 13.51/3 = 2.381…
  3. Median ≈ 2.38
9.For f(x) = (4 − x) on 0 ≤ x ≤ 4, E(X) = (answer space).

Answer:

E(X) = ∫ from 0 to 4 of (4x − x2) dx = [2x2 − ] from 0 to 4 = (32 − ) = × = .

Part C: Cumulative distribution functions

The cumulative distribution function (cdf) is F(x) = P(X ≤ x), the area under f to the left of x. It rises from 0 to 1, and f(x) = F′(x).

With a cdf, probabilities need no integration: P(a < X ≤ b) = F(b) − F(a), and the median solves F(m) = .

Worked example. f(x) = 2x on 0 ≤ x ≤ 1 has F(x) = x2 for 0 ≤ x ≤ 1. So P(0.5 < X < 0.8) = 0.64 − 0.25 = 0.39.

10.For f(x) = on 0 ≤ x ≤ 3, find the cumulative distribution function F(x) for 0 ≤ x ≤ 3, and evaluate F(1.5).

Answer: 0.125

  1. F(x) = ∫ from 0 to x of dt =
  2. F(1.5) = = 0.125
  3. So P(X ≤ 1.5) = 0.125.
11.A random variable has cumulative distribution function F(x) = for 0 ≤ x ≤ 4. Find its density function and P(1 < X < 3).

Answer:

  1. f(x) = F′(x) = = for 0 ≤ x ≤ 4
  2. P(1 < X < 3) = F(3) − F(1) = − = = [frac 1/2]
12.For the distribution with F(x) = on 0 ≤ x ≤ 4, find the median, correct to 2 decimal places.

Answer: 2.83

  1. F(m) = : = , so m2 = 8
  2. m = √8 = 2√2 ≈ 2.83

Part D: Uniform and normal distributions

Uniform distribution on a ≤ x ≤ b: f(x) = , E(X) = , Var(X) = . Probabilities are rectangle areas.

Normal distribution N(μ, σ2): standardise with z = [frac x − μ/σ] and use a table or calculator for P(Z < z). To work backwards from a probability, find z first, then x = μ + zσ.

Worked example. X ~ N(100, 152). P(X < 115) = P(Z < 1) = 0.8413, and the value with 2.5% above it is 100 + 1.96 × 15 = 129.4.

13.A waiting time X is uniformly distributed between 0 and 12 minutes. P(X > 9) = (answer space) and E(X) = (answer space) minutes.

Answer: 0.25, 6

f(x) = on 0 ≤ x ≤ 12. P(X > 9) = (12 − 9) × = 0.25. E(X) = = 6. (The standard deviation is = √12 ≈ 3.46 minutes.)

14.X is uniformly distributed on 2 ≤ x ≤ 10. What is Var(X)?
  1. A
  2. B4
  3. C8
  4. D

Answer:

For a uniform distribution on [a, b], Var(X) = = = ≈ 5.33.

15.The lengths of bolts are normally distributed with mean 50 mm and standard deviation 4 mm. Find the probability that a bolt is longer than 56 mm, correct to 4 decimal places.

Answer: 0.0668

  1. z = = 1.5
  2. P(X > 56) = P(Z > 1.5) = 1 − 0.9332
  3. = 0.0668
16.For the bolts in the previous question, find the length exceeded by only 5% of bolts, correct to 1 decimal place.

Answer: 56.6 mm

  1. Find z with P(Z > z) = 0.05, that is P(Z < z) = 0.95: z = 1.645
  2. x = μ + zσ = 50 + 1.645 × 4 = 56.58
  3. = 56.6 mm

Part E: Extended problem

The lifetime T, in years, of an electronic component has probability density function f(t) = t(4 − t) for 0 ≤ t ≤ 4, and 0 otherwise.

(a) Show that f is a valid probability density function.

(b) Find the probability that a component fails within its first year.

(c) Find the expected lifetime.

(d) Find the standard deviation of the lifetime, correct to 3 decimal places.

(e) The maker replaces, free of charge, any component that fails within 1 year. Of 2000 components sold, how many would you expect to be replaced?

(f) A component is still working after 1 year. Find the probability that it lasts more than 3 years.

17.Answer parts (a) to (f), showing full working.[12 marks]

Model answer: (a) f(t) = t(4 − t) ≥ 0 on 0 ≤ t ≤ 4, and ∫ from 0 to 4 of (4t − t2) dt = [2t2 − ] from 0 to 4 = (32 − ) = × = 1. (b) P(T < 1) = (2 − ) = × = [frac 5/32] ≈ 0.156. (c) The graph is symmetric about t = 2, so E(T) = 2 years (check: ∫ (4t2 − t3) dt = (256/3 − 64) = 2). (d) E(T2) = ∫ from 0 to 4 of (4t3 − t4) dt = (256 − 204.8) = 4.8, so Var(T) = 4.8 − 4 = 0.8 and σ = √0.8 ≈ 0.894 years. (e) 2000 × = 312.5, so about 313 components. (f) By symmetry P(T > 3) = and P(T > 1) = , so P(T > 3 | T > 1) = ÷ = [frac 5/27] ≈ 0.185.

Success criteria:

  • (a) shows f(t) ≥ 0 and that the total area is 1 (2 marks)
  • (b) ≈ 0.156 (2 marks)
  • (c) E(T) = 2, by symmetry or integration (1 mark)
  • (d) E(T2) = 4.8, Var = 0.8, σ ≈ 0.894 (3 marks)
  • (e) about 313 components (1 mark)
  • (f) conditional probability ≈ 0.185 (3 marks)
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