Continuous Random Variables and PDFs — Full Worked Solutions

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Full Worked Solutions

1. Question: Show that f(x) = 3x² on [0, 1] is a valid pdf by verifying both conditions. Then find k if instead f(x) = kx² on [0, 2].

   Verifying f(x) = 3x² on [0, 1]

   Condition 1 — Non-negativity:
   For x ∈ [0, 1]: x² ≥ 0, so 3x² ≥ 0. ✓

   Condition 2 — Total area = 1:
   ∫₀¹ 3x² dx = [x³]₀¹ = 1 − 0 = 1 ✓

   Both conditions satisfied; f(x) = 3x² is a valid pdf.

   Find k for f(x) = kx² on [0, 2]

   Set ∫₀² kx² dx = 1:
   k[x³/3]₀² = 1
   k × 8/3 = 1
   k = 3/8

   Check: f(x) = (3/8)x² ≥ 0 on [0, 2] ✓

2. Question: Given f(x) = (3/4)(1 − x²) on [−1, 1], find P(0 ≤ X ≤ 0.5).

   Setting Up the Integral

   P(0 ≤ X ≤ 0.5) = ∫₀^0.5 (3/4)(1 − x²) dx

   Evaluating

   = (3/4)[x − x³/3]₀^0.5
   = (3/4)[(0.5 − (0.5)³/3) − 0]
   = (3/4)[0.5 − 0.125/3]
   = (3/4)[0.5 − 0.04167]
   = (3/4) × 0.45833
   = 0.344 (to 3 d.p.)

3. Question: For f(x) = 3x² on [0, 1], find the CDF F(x) for x ∈ [0, 1].

   Deriving the CDF

   The CDF is obtained by integrating the pdf from the lower bound:

   F(x) = ∫₀^x 3t² dt = [t³]₀^x = x³   for x ∈ [0, 1]

   Verification

   • F(0) = 0³ = 0 ✓ (no probability below the lower bound)
   • F(1) = 1³ = 1 ✓ (all probability is accumulated by the upper bound)

4. Question: Using f(x) = 3x² on [0, 1] (with CDF F(x) = x³), find: (a) P(X > 0.7)   (b) The median of the distribution.

   (a) P(X > 0.7)

   Using the complement rule:
   P(X > 0.7) = 1 − F(0.7) = 1 − (0.7)³ = 1 − 0.343 = 0.657

   (b) Finding the Median

   The median m satisfies F(m) = 0.5:
   m³ = 0.5
   m = (0.5)^1/3 = 0.794 (to 3 d.p.)

   Check: m ∈ [0, 1] ✓

5. Question: Given f(x) = k(4 − x) on [0, 4]: (a) Find k   (b) Find P(1 ≤ X ≤ 3)   (c) Find P(X > 2.5)

   (a) Find k

   ∫₀⁴ k(4 − x) dx = 1
   k[4x − x²/2]₀⁴ = 1
   k[(16 − 8) − 0] = 1
   8k = 1
   k = 1/8

   (b) P(1 ≤ X ≤ 3)

   = ∫₁³ (4 − x)/8 dx = (1/8)[4x − x²/2]₁³
   = (1/8)[(12 − 9/2) − (4 − 1/2)]
   = (1/8)[(12 − 4.5) − (4 − 0.5)]
   = (1/8)[7.5 − 3.5]
   = 4/8 = 0.5

   (c) P(X > 2.5)

   = ∫_2.5⁴ (4 − x)/8 dx = (1/8)[4x − x²/2]_2.5⁴
   = (1/8)[(16 − 8) − (10 − 3.125)]
   = (1/8)[8 − 6.875]
   = 1.125/8 = 0.141 (to 3 d.p.)

6. Question: A piecewise pdf is defined by f(x) = 2x for 0 ≤ x ≤ 1, f(x) = 0 otherwise. Verify this is a valid pdf, then find the CDF F(x) and use it to find P(0.4 ≤ X ≤ 0.8).

   Verification

   Non-negativity: f(x) = 2x ≥ 0 for x ∈ [0, 1] ✓
   Total area: ∫₀¹ 2x dx = [x²]₀¹ = 1 ✓

   CDF F(x)

   For 0 ≤ x ≤ 1:
   F(x) = ∫₀^x 2t dt = [t²]₀^x = x²

   P(0.4 ≤ X ≤ 0.8)

   = F(0.8) − F(0.4)
   = (0.8)² − (0.4)²
   = 0.64 − 0.16 = 0.48

7. Question: Find the constant k so that f(x) = kx(1 − x) is a valid pdf on [0, 1]. Then find the median of this distribution.

   Find k

   ∫₀¹ kx(1 − x) dx = 1
   k ∫₀¹ (x − x²) dx = 1
   k[x²/2 − x³/3]₀¹ = 1
   k(1/2 − 1/3) = 1
   k × 1/6 = 1
   k = 6

   Find the Median

   Note that f(x) = 6x(1 − x) is symmetric about x = 1/2 (since f(1 − x) = 6(1 − x)x = f(x)). Therefore, by symmetry, the median is m = 1/2.

   Verification: ∫₀^0.5 6x(1 − x) dx = 6[x²/2 − x³/3]₀^0.5
   = 6(0.125 − 0.04167) = 6 × 0.08333 = 0.5 ✓

8. Question: The pdf of X is f(x) = c/x² for x ∈ [1, 5]. (a) Find c   (b) Find P(X > 3)   (c) Find the CDF F(x)

   (a) Find c

   ∫₁⁵ c/x² dx = 1
   c[−1/x]₁⁵ = 1
   c(−1/5 − (−1)) = 1
   c(1 − 1/5) = 1
   c × 4/5 = 1
   c = 5/4

   (b) P(X > 3)

   = ∫₃⁵ (5/4)/x² dx = (5/4)[−1/x]₃⁵
   = (5/4)(−1/5 − (−1/3))
   = (5/4)(1/3 − 1/5)
   = (5/4)(2/15)
   = 10/60 = 1/6 ≈ 0.167

   (c) CDF F(x)

   F(x) = ∫₁^x (5/4)/t² dt = (5/4)[−1/t]₁^x
   = (5/4)(1 − 1/x)
   = 5(x − 1)/(4x)   for x ∈ [1, 5]

   Check: F(1) = 0 ✓, F(5) = 5(4)/(20) = 1 ✓

9. Question: A random variable X has pdf f(x) = ax + b on [0, 2], where f(0) = 0.2 and f(2) = 0.8. (a) Find a and b   (b) Verify that f(x) is a valid pdf   (c) Find P(X ≤ 1) and the median.

   (a) Find a and b

   Using the given boundary values:
   f(0) = a(0) + b = b = 0.2, so b = 0.2
   f(2) = 2a + 0.2 = 0.8, so 2a = 0.6, a = 0.3
   Therefore: f(x) = 0.3x + 0.2

   (b) Verify f(x) is a Valid PDF

   Non-negativity: f(x) = 0.3x + 0.2 ≥ 0.2 > 0 on [0, 2] ✓
   Total area:
   ∫₀² (0.3x + 0.2) dx = [0.15x² + 0.2x]₀²
   = (0.15 × 4 + 0.4) − 0 = 0.6 + 0.4 = 1 ✓

   (c) P(X ≤ 1) and Median

   P(X ≤ 1):
   = ∫₀¹ (0.3x + 0.2) dx = [0.15x² + 0.2x]₀¹ = 0.15 + 0.2 = 0.35

   Median m:
   [0.15x² + 0.2x]₀^m = 0.5
   0.15m² + 0.2m = 0.5
   Multiply by 20: 3m² + 4m − 10 = 0
   Using the quadratic formula:
   m = (−4 + √(16 + 120)) / 6 = (−4 + √136) / 6
   √136 ≈ 11.662
   m ≈ 7.662 / 6 ≈ 1.277 (taking the positive root since m ∈ [0, 2])

10. Question: A machine produces components with a fault occurring at position X (in mm from one end) modelled by f(x) = k(6x − x²) on [0, 6]. (a) Find k   (b) Find P(2 ≤ X ≤ 4)   (c) A component is rejected if the fault position is less than 1 mm or greater than 5 mm from the end. Find the probability that a component is rejected.

    (a) Find k

    ∫₀⁶ k(6x − x²) dx = 1
    k[3x² − x³/3]₀⁶ = 1
    k[(108 − 72) − 0] = 1
    36k = 1
    k = 1/36

    (b) P(2 ≤ X ≤ 4)

    = (1/36) ∫₂⁴ (6x − x²) dx
    = (1/36)[3x² − x³/3]₂⁴
    = (1/36)[(48 − 64/3) − (12 − 8/3)]
    = (1/36)[(48 − 21.33) − (12 − 2.67)]
    = (1/36)[26.67 − 9.33]
    = (1/36)(17.33)
    ≈ 0.481

    (c) P(Rejected)

    P(rejected) = P(X < 1) + P(X > 5)

    P(X < 1):
    = (1/36)[3x² − x³/3]₀¹
    = (1/36)(3 − 1/3)
    = (1/36)(8/3) = 8/108 ≈ 0.0741

    P(X > 5): By symmetry of f(x) = (1/36)(6x − x²) about x = 3 (since f(3 + t) = f(3 − t)):
    P(X > 5) = P(X < 1) ≈ 0.0741

    P(rejected) = 0.0741 + 0.0741 ≈ 0.148