---
title: "Normal: mean and SD from rod diameters"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:rod-diameter-mean-sd-normal"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/rod-diameter-mean-sd-normal
site: BCT Engineering
---

# Normal: mean and SD from rod diameters

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2076 Chaitra, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/rod-diameter-mean-sd-normal
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

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.

## 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.

## Answer

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**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. To find mu and sigma we:
   A. Convert both percentages to z-values and solve two equations
   B. Average 35 and 63
   C. Use one equation only
   D. Take sigma = 63 - 35
   Answer: A. Two conditions, two unknowns. (Key point: Approach)

2. P(X < 35) = 0.07 gives z of:
   A. +1.48
   B. -0.07
   C. About -1.48 (negative)
   D. +0.07
   Answer: C. 7% is below the mean so z is negative. (Key point: z-value)

3. The two equations are:
   A. mu + 1.48 sigma = 35 and mu - 1.23 sigma = 63
   B. mu - 1.48 sigma = 63 and mu + 1.23 sigma = 35
   C. 1.48 mu = 35
   D. mu - 1.48 sigma = 35 and mu + 1.23 sigma = 63
   Answer: D. X = mu + z sigma. (Key point: Equations)

4. The solution is:
   A. mu = 49, sigma = 14
   B. mu = 50.29, sigma = 10.33
   C. mu = 10.33, sigma = 50.29
   D. mu = 35, sigma = 63
   Answer: B. Subtracting: 2.71 sigma = 28. (Key point: Result)

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