Solutions — The Normal Distribution

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1. Show Solution
   X ~ N(20, 4) means μ = 20, σ² = 4, so σ = 2
   (a) μ = 20, σ = 2
   (b) P(X < 22) = normCdf(−10⁹⁹, 22, 20, 2) ≈ 0.8413
   (Z = (22−20)/2 = 1; P(Z < 1) = 0.8413)
   (c) P(X > 20) = 0.5 by symmetry (20 is the mean)
2. Show Solution
   X ~ N(50, 100) so μ = 50, σ = 10
   (a) 40 = μ − σ and 60 = μ + σ
   P(40 < X < 60) = P(μ−σ < X < μ+σ) ≈ 0.6827 (68.27%)
   (b) 70 = μ + 2σ
   P(X > 70) = [1 − P(μ−2σ < X < μ+2σ)] / 2 = (1 − 0.9545)/2 ≈ 0.0228 (2.28%)
3. Show Solution
   X ~ N(30, 25), so μ = 30, σ = 5
   Formula: Z = (x − μ) / σ = (x − 30) / 5
   (a) Z = (35 − 30)/5 = 1.0
   (b) Z = (22 − 30)/5 = −1.6
   (c) Z = (30 − 30)/5 = 0 (the mean always gives Z = 0)
4. Show Solution
   Z ~ N(0, 1)
   (a) P(Z < 1.5) = normCdf(−10⁹⁹, 1.5, 0, 1) ≈ 0.9332
   (b) P(Z > −0.8) = 1 − P(Z < −0.8)
   P(Z < −0.8) ≈ 0.2119, so P(Z > −0.8) ≈ 0.7881
   (Or: by symmetry P(Z > −0.8) = P(Z < 0.8) ≈ 0.7881)
   (c) P(−1 < Z < 2) = normCdf(−1, 2, 0, 1) ≈ 0.8186
5. Show Solution
   X ~ N(172, 36) so μ = 172, σ = 6
   (a) P(X < 165) = normCdf(−10⁹⁹, 165, 172, 6)
   Z = (165−172)/6 = −7/6 ≈ −1.167
   P(X < 165) ≈ 0.1216
   (b) P(X > 180) = 1 − normCdf(−10⁹⁹, 180, 172, 6)
   Z = (180−172)/6 = 8/6 ≈ 1.333
   P(X > 180) ≈ 0.0912
   (c) P(166 < X < 178) = normCdf(166, 178, 172, 6)
   Z₁ = −1, Z₂ = 1; P(−1 < Z < 1) ≈ 0.6827
6. Show Solution
   X ~ N(45, 64) so μ = 45, σ = 8
   Z₁ = (38 − 45)/8 = −7/8 = −0.875
   Z₂ = (55 − 45)/8 = 10/8 = 1.25
   P(38 < X < 55) = P(−0.875 < Z < 1.25)
   = normCdf(38, 55, 45, 8) ≈ 0.7029
7. Show Solution
   The normal distribution is perfectly symmetric about its mean μ. This means the curve is a mirror image on either side of μ. Since the total area = 1, and the left half is a perfect mirror of the right half, each half has area = 1/2 = 0.5.
   
   Therefore P(X < μ) = 0.5 for any normal distribution, regardless of the values of μ and σ.
   
   [Diagram: Bell curve centred at μ, with left half shaded, area = 0.5 labelled.]
8. Show Solution
   P(X < 70) = 0.8413 ⇒ this is approximately P(Z < 1) = 0.8413
   So the value x = 70 corresponds to a Z-score of 1.
   Using Z = (x − μ) / σ:
   1 = (70 − μ) / 5
   5 = 70 − μ
   μ = 65
9. Show Solution
   X ~ N(280, 400) so μ = 280, σ = 20
   (a) P(260 < X < 320)
   260 = μ − σ, 320 = μ + 2σ
   P(260 < X < 320) = normCdf(260, 320, 280, 20) ≈ 0.8186
   (b) P(X < 250) = normCdf(−10⁹⁹, 250, 280, 20)
   Z = (250−280)/20 = −1.5
   P(X < 250) ≈ 0.0668 = 6.68%
   (c) P(X > 310) = 1 − normCdf(−10⁹⁹, 310, 280, 20)
   Z = (310−280)/20 = 1.5
   P(X > 310) ≈ 0.0668
   Expected number = 500 × 0.0668 ≈ 33 avocados
10. Show Solution
    Factory A: X ~ N(1200, 10000), σ_A = 100
    Factory B: Y ~ N(1150, 3600), σ_B = 60
    (a) P(X > 1300): Z = (1300−1200)/100 = 1; P(X > 1300) = 1 − P(Z < 1) ≈ 0.1587
    P(Y > 1300): Z = (1300−1150)/60 = 2.5; P(Y > 1300) = 1 − P(Z < 2.5) ≈ 0.0062
    (b) Factory A is more likely to produce a bulb lasting over 1300 hours (probability 0.1587 vs 0.0062), despite having a higher mean. Factory A’s larger spread (σ = 100) means more values in the upper tail.
    (c) P(Y < 1000): Z = (1000−1150)/60 = −2.5; P(Y < 1000) ≈ 0.0062
    Only 0.62% of Factory B bulbs last under 1000 hours, suggesting Factory B is very consistent (small spread) even though its mean is slightly lower.