---
title: "Breakdown voltage: normal probabilities (1.5 V)"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:bv-39-42-le43-ge39"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bv-39-42-le43-ge39
site: BCT Engineering
---

# Breakdown voltage: normal probabilities (1.5 V)

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2075 Chaitra, 5 marks; 2075 Asoj, 6 marks; 2069 Chaitra, 6 marks (asked 3 times)
- Video: [Normal Distribution EXPLAINED with Examples](https://www.youtube.com/watch?v=xI9ZHGOSaCg) — Ace Tutors
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bv-39-42-le43-ge39
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

The breakdown voltage X of a randomly chosen diode of a particular type is known to be normally distributed with mean 40 and standard deviation 1.5 volts. What is the probability that the breakdown voltage will be (a) between 39 and 42 volts; (b) at most 43 volts; (c) at least 39 volts.

## Short answer

Standardize with Z=(X-40)/1.5 and use the normal table. The probabilities are (a) 0.6568, (b) 0.9772, and (c) 0.7486.

## Answer

X ~ N(μ = 40, σ = 1.5). Standardise with Z = (X − μ)/σ and use the standard normal table Φ(z) = P(Z ≤ z).

#### (a) P(39 < X < 42)
z₁ = (39 − 40)/1.5 = −0.67, z₂ = (42 − 40)/1.5 = 1.33

P = Φ(1.33) − Φ(−0.67) = 0.9082 − 0.2514 = **0.6568**

#### (b) P(X ≤ 43)
z = (43 − 40)/1.5 = 2.00

P = Φ(2.00) = **0.9772**

#### (c) P(X ≥ 39)
z = (39 − 40)/1.5 = −0.67

P = 1 − Φ(−0.67) = 1 − 0.2514 = **0.7486**

(Values are from standard normal tables with z rounded to 2 decimals; exact computation gives 0.6563, 0.9772, 0.7475.)

**Answer: (a) 0.6568, (b) 0.9772, (c) 0.7486**

## Check yourself

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

1. To use the table, X is converted by:
   A. Z = (X - 1.5)/40
   B. Z = X/40
   C. Z = (X - 40) x 1.5
   D. Z = (X - 40)/1.5
   Answer: D. Z = (X - mu)/sigma. (Key point: Standardising)

2. P(39 < X < 42) is:
   A. Phi(1.33) + Phi(-0.67)
   B. Phi(1.33) - Phi(-0.67)
   C. Phi(42) - Phi(39)
   D. 1 - Phi(1.33)
   Answer: B. Subtract the lower cumulative area from the upper. (Key point: Between)

3. z for 43 is:
   A. 2.00, giving P = 0.9772
   B. 3.00
   C. 1.50
   D. 0.67
   Answer: A. (43 - 40)/1.5 = 2. (Key point: At most 43)

4. P(X >= 39) is:
   A. Phi(-0.67) = 0.2514
   B. 1 - Phi(0.67)=0.2514
   C. 1 - Phi(-0.67) = 0.7486
   D. Phi(1.33)
   Answer: C. Tables give lower-tail area, so take the complement. (Key point: At least 39)

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- [Breakdown voltage: normal, variance 2.25](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abv-39-42-lt44-gt43) — Probability and Statistics, chapter 3, asked 3×
- [Breakdown voltage: three ranges](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abv-39-42-40-43-lt44) — Probability and Statistics, chapter 3, asked 3×

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