Practice Maths

Topic Review — Continuous Random Variables and Normal Distribution — Solutions

← Continuous RV and Normal Distribution

This review covers all lessons in this topic: continuous random variables and PDFs, expected value and variance of continuous RVs, the normal distribution, and using the normal distribution (inverse normal, finding parameters). Click each answer to reveal the worked solution.

Review Questions

  1. State two conditions that a probability density function (PDF) f(x) must satisfy.
    1. f(x) ≥ 0 for all x (non-negative)
    2. ∫−∞∞ f(x) dx = 1 (total area = 1)
  2. A continuous random variable X has PDF f(x) = kx for 0 ≤ x ≤ 4 and 0 elsewhere. Find k.
    ∫04 kx dx = 1
    k[x²/2]04 = 1
    k(8) = 1
    k = 1/8
  3. For the PDF f(x) = 1/8 · x on [0, 4], find P(1 < X < 3).
    P(1 < X < 3) = ∫13 (x/8) dx = [x²/16]13 = 9/16 − 1/16 = 8/16 = 1/2
  4. A continuous random variable has PDF f(x) = 3x² for 0 ≤ x ≤ 1. Find E(X).
    E(X) = ∫01 x ċ 3x² dx = ∫01 3x³ dx = [3x4/4]01 = 3/4
  5. For the PDF f(x) = 3x² on [0,1], find E(X²) and hence Var(X).
    E(X²) = ∫01 x² ċ 3x² dx = [3x5/5]01 = 3/5
    Var(X) = E(X²) − [E(X)]² = 3/5 − (3/4)² = 3/5 − 9/16
    = 48/80 − 45/80 = 3/80
  6. X ~ N(80, 25). Find: (a) μ and σ   (b) P(X < 85) using CAS   (c) P(70 < X < 90)
    (a) μ = 80, σ² = 25, so σ = 5
    (b) P(X < 85) = normCdf(−1099, 85, 80, 5) ≈ 0.8413
    (Z = (85−80)/5 = 1; P(Z < 1) = 0.8413)
    (c) P(70 < X < 90): Z1 = −2, Z2 = 2
    P = P(−2 < Z < 2) ≈ 0.9545
  7. Use the 68–95–99.7 rule to answer: X ~ N(200, 400) (so σ = 20). What percentage of values are greater than 240?
    240 = 200 + 2(20) = μ + 2σ
    P(μ−2σ < X < μ+2σ) = 0.9545
    By symmetry, P(X > μ+2σ) = (1−0.9545)/2 = 0.02275
    About 2.28% of values are greater than 240.
  8. The time (in minutes) students spend on homework daily is normally distributed with mean 45 min and standard deviation 10 min. Find the probability that a randomly selected student spends less than 30 minutes on homework.
    X ~ N(45, 100), σ = 10
    Z = (30 − 45)/10 = −1.5
    P(X < 30) = normCdf(−1099, 30, 45, 10) ≈ 0.0668
  9. X ~ N(60, 16). Find the value c such that P(X > c) = 0.1.
    P(X > c) = 0.1 ⇒ P(X < c) = 0.9
    c = invNorm(0.9, 60, 4)
    Z = invNorm(0.9, 0, 1) ≈ 1.2816
    c = 60 + 1.2816 × 4 ≈ 65.13
  10. X ~ N(μ, 49). Given P(X < 30) = 0.6, find μ.
    σ = 7
    P(X < 30) = 0.6 ⇒ Z = invNorm(0.6, 0, 1) ≈ 0.2533
    30 = μ + 0.2533 × 7
    30 = μ + 1.773
    μ ≈ 28.23
  11. X ~ N(50, σ²). Given P(X < 60) = 0.8, find σ.
    Z = invNorm(0.8, 0, 1) ≈ 0.8416
    0.8416 = (60 − 50) / σ
    σ = 10 / 0.8416 ≈ 11.88
  12. Weights of apples are normally distributed. P(X < 150) = 0.2 and P(X < 200) = 0.9. Find the mean and standard deviation.
    Z1 = invNorm(0.2, 0, 1) ≈ −0.8416
    Z2 = invNorm(0.9, 0, 1) ≈ 1.2816
    Equations: 150 = μ − 0.8416σ and 200 = μ + 1.2816σ
    Subtracting: 50 = 2.1232σ ⇒ σ ≈ 23.55
    μ = 150 + 0.8416 × 23.55 ≈ 169.82
    So μ ≈ 169.8 g, σ ≈ 23.6 g
  13. A normal distribution has the property that P(X < 42) = P(X > 58). What is the mean? Explain your reasoning.
    P(X < 42) = P(X > 58) means the values 42 and 58 are symmetric about the mean.
    μ = (42 + 58)/2 = 50
    This works because the normal distribution is symmetric about μ, so values equidistant from the mean have equal tail probabilities.
  14. A continuous random variable X has PDF f(x) = c(4 − x²) for −2 ≤ x ≤ 2 and 0 elsewhere.
    • (a) Find c
    • (b) Find E(X) and explain the result using symmetry
    • (c) Find P(0 < X < 1)
    (a) ∫−22 c(4−x²) dx = 1
    c[4x − x³/3]−22 = 1
    c[(8−8/3) − (−8+8/3)] = c[16−16/3] = c(32/3) = 1
    c = 3/32
    (b) E(X) = ∫−22 x · (3/32)(4−x²) dx
    The integrand x(4−x²) = 4x − x³ is an odd function (symmetric domain [−2,2]).
    E(X) = 0. The PDF is symmetric about x = 0, so the mean is 0.
    (c) P(0 < X < 1) = ∫01 (3/32)(4−x²) dx = (3/32)[4x−x³/3]01
    = (3/32)(4−1/3) = (3/32)(11/3) = 11/32 ≈ 0.344
  15. Scores on a standardised test are normally distributed with μ = 500 and σ = 100. A scholarship requires a score in the top 5%.
    • (a) Find the minimum score required for the scholarship.
    • (b) What percentage of students score between 400 and 650?
    • (c) If 2000 students sit the test, how many are expected to score above 700?
    (a) P(X > x) = 0.05 ⇒ P(X < x) = 0.95
    x = invNorm(0.95, 500, 100) ≈ 664.5
    (b) P(400 < X < 650) = normCdf(400, 650, 500, 100)
    Z1 = −1, Z2 = 1.5
    P ≈ 0.8413 − 0.1587 + [P(Z < 1.5) − 0.5] = normCdf(400, 650, 500, 100) ≈ 0.7745 (77.45%)
    (c) P(X > 700): Z = (700−500)/100 = 2
    P(X > 700) = P(Z > 2) ≈ 0.0228
    Expected number = 2000 × 0.0228 ≈ 46 students