Solutions — Using the Normal Distribution

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1. Show Solution
   X ~ N(40, 9), μ = 40, σ = 3
   (a) P(X < a) = 0.95
   a = invNorm(0.95, 40, 3) ≈ 44.93
   (Z = 1.6449; a = 40 + 1.6449 × 3 ≈ 44.93)
   (b) P(X > b) = 0.2 ⇒ P(X < b) = 0.8
   b = invNorm(0.8, 40, 3) ≈ 42.52
2. Show Solution
   X ~ N(50, 16), μ = 50, σ = 4
   Q₁: P(X < Q₁) = 0.25
   Q₁ = invNorm(0.25, 50, 4) ≈ 47.30
   Q₃: P(X < Q₃) = 0.75
   Q₃ = invNorm(0.75, 50, 4) ≈ 52.70
   (Symmetrically placed about the mean, as expected.)
3. Show Solution
   X ~ N(μ, 25), σ = 5
   P(X < 45) = 0.8413 ⇒ Z ≈ 1 (since P(Z < 1) = 0.8413)
   Using Z = (x − μ)/σ:
   1 = (45 − μ) / 5
   5 = 45 − μ
   μ = 40
4. Show Solution
   X ~ N(70, σ²)
   P(X < 80) = 0.9772 ⇒ Z ≈ 2 (since P(Z < 2) = 0.9772)
   Using Z = (x − μ)/σ:
   2 = (80 − 70) / σ
   2σ = 10
   σ = 5
5. Show Solution
   By symmetry of the normal distribution, the mean lies midway between the 10th and 90th percentiles:
   μ = (490 + 510)/2 = 500 g
   
   P(X < 510) = 0.9 ⇒ Z = invNorm(0.9, 0, 1) ≈ 1.2816
   1.2816 = (510 − 500) / σ
   σ = 10 / 1.2816 ≈ 7.80 g
6. Show Solution
   P(X < 25) = 0.3 ⇒ Z₁ = invNorm(0.3, 0, 1) ≈ −0.5244
   P(X < 40) = 0.85 ⇒ Z₂ = invNorm(0.85, 0, 1) ≈ 1.0364
   Equations:
   25 = μ − 0.5244σ … (i)
   40 = μ + 1.0364σ … (ii)
   (ii) − (i): 15 = 1.5608σ ⇒ σ = 15/1.5608 ≈ 9.61
   From (i): μ = 25 + 0.5244 × 9.61 ≈ 30.04
7. Show Solution
   X ~ N(10, 0.04), σ = 0.2 mm
   P(9.5 < X < 10.5) = normCdf(9.5, 10.5, 10, 0.2)
   Z₁ = (9.5 − 10)/0.2 = −2.5
   Z₂ = (10.5 − 10)/0.2 = 2.5
   P(−2.5 < Z < 2.5) ≈ 0.9876 (98.76%)
8. Show Solution
   X ~ N(100, 225), σ = 15
   Top 2% means P(X > x) = 0.02, i.e., P(X < x) = 0.98
   x = invNorm(0.98, 100, 15) ≈ 130.8
   (Z = invNorm(0.98, 0, 1) ≈ 2.054; x = 100 + 2.054 × 15 ≈ 130.8)
   Minimum IQ for Mensa is approximately 131.
9. Show Solution
   A ~ N(50, 4), σ_A = 2;   B ~ N(80, 9), σ_B = 3
   (a) P(A < 48): Z = (48−50)/2 = −1
   P(A < 48) = P(Z < −1) ≈ 0.1587
   (b) P(B > b) = 0.95 ⇒ P(B < b) = 0.05
   b = invNorm(0.05, 80, 3) ≈ 80 − 1.6449 × 3 ≈ 75.07 mm
   95% of Type B components are longer than 75.07 mm.
   (c) Type A: P(47 < A < 53)
   Z₁ = −1.5, Z₂ = 1.5
   P = normCdf(47, 53, 50, 2) ≈ 0.8664 ≈ 86.64%
   Type B: P(75 < B < 85)
   Z₁ = −1.667, Z₂ = 1.667
   P = normCdf(75, 85, 80, 3) ≈ 0.9044 ≈ 90.44%
10. Show Solution
    (a) P(X < 160) = 0.20 ⇒ Z₁ = invNorm(0.2, 0, 1) ≈ −0.8416
    Equation 1: 160 = μ − 0.8416σ
    P(X > 180) = 0.15 ⇒ P(X < 180) = 0.85 ⇒ Z₂ = invNorm(0.85, 0, 1) ≈ 1.0364
    Equation 2: 180 = μ + 1.0364σ
    (b) (Eq2) − (Eq1): 20 = (1.0364 + 0.8416)σ = 1.8780σ
    σ = 20/1.8780 ≈ 10.65
    From Eq1: μ = 160 + 0.8416 × 10.65 ≈ 168.96
    (c) P(165 < X < 175) = normCdf(165, 175, 168.96, 10.65)
    ≈ 0.3673 (36.73%)