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Normal: 31% under 45, 8% over 64

Short answer

From P(X<45)=0.31, z=-0.5; P(X>64)=0.08, z=1.4. Equations: μ−0.5σ=45, μ+1.4σ=64. Solving: σ=10, μ=50.

In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the mean and standard deviation of distribution. (Given, Z0.42 = 1.4, Z0.19 = 0.5)

Let X ~ N(μ, σ²). Given: P(X < 45) = 0.31 and P(X > 64) = 0.08.

Z-values

  • P(X < 45) = 0.31 lies below the mean, so the area between 45 and μ is 0.50 − 0.31 = 0.19. Given Z₀.₁₉ = 0.5, the z-value is −0.5.
  • P(X > 64) = 0.08, so the area between μ and 64 is 0.50 − 0.08 = 0.42. Given Z₀.₄₂ = 1.4, the z-value is +1.4.

Equations

(45 − μ)/σ = −0.5 → μ − 0.5σ = 45 ... (1) (64 − μ)/σ = 1.4 → μ + 1.4σ = 64 ... (2)

Subtract (1) from (2): 1.9σ = 19 → σ = 10

From (1): μ = 45 + 0.5 × 10 = 50

(Check: (64 − 50)/10 = 1.4.)

Answer: mean = 50, standard deviation = 10

✓ Answer checked by Claude Sonnet 5.5

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