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Normal: mean and SD from rod diameters

Short answer

Use z-values -1.48 for 0.07 and 1.23 for 0.89. Solving mu - 1.48 sigma = 35 and mu + 1.23 sigma = 63 gives mean about 50.29 mm and standard deviation about 10.33 mm.

If inner diameter of a rod follow Normal Distribution. If 7% of the rod has inner diameter less than 35 mm and 89% of rod has inner diameter fewer than 63 mm. Find the mean and the standard deviation of diameter of rods.

Let X ~ N(μ, σ²) be the inner diameter. Two conditions are given:

  • P(X < 35) = 0.07
  • P(X < 63) = 0.89

Find the z-values

  • P(Z < z₁) = 0.07 gives z₁ = −1.48 (table: Φ(−1.48) = 0.0694; exact z = −1.476).
  • P(Z < z₂) = 0.89 gives z₂ = +1.23 (table: Φ(1.23) = 0.8907).

Equations

(35 − μ)/σ = −1.48 → μ − 1.48σ = 35 ... (1) (63 − μ)/σ = 1.23 → μ + 1.23σ = 63 ... (2)

Subtract (1) from (2): 2.71σ = 28 → σ = 28/2.71 = 10.33

From (2): μ = 63 − 1.23 × 10.33 = 63 − 12.71 = 50.29

(Check with (1): 50.29 − 1.48 × 10.33 = 35.00.)

Answer: mean ≈ 50.29 mm, standard deviation ≈ 10.33 mm

✓ Answer checked by Claude Sonnet 5.5

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