---
title: "Breakdown voltage: normal, variance 2.25"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:bv-39-42-lt44-gt43"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bv-39-42-lt44-gt43
site: BCT Engineering
---

# Breakdown voltage: normal, variance 2.25

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2071 Chaitra, 5 marks; 2074 Asoj, 5 marks (asked 2 times)
- Video: [Z Scores and Normal Distributions (Example Problems)](https://www.youtube.com/watch?v=rkdbHQPUA1Y) — Mario's Math Tutoring
- 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-lt44-gt43
- 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 volts and variance 2.25 volts. What is the probability that the breakdown voltage will be
i) Between 39 and 42 volts
ii) Less than 44 volts
iii) More than 43 volts

## Short answer

Standardize with sigma=1.5. The probabilities are (i) 0.6568, (ii) 0.9962, and (iii) 0.0228.

## Answer

X ~ N(μ = 40, σ² = 2.25), so σ = √2.25 = 1.5. Use Z = (X − 40)/1.5.

#### i) 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**

#### ii) P(X < 44)
z = (44 − 40)/1.5 = 2.67

P = Φ(2.67) = **0.9962**

#### iii) P(X > 43)
z = (43 − 40)/1.5 = 2.00

P = 1 − Φ(2.00) = 1 − 0.9772 = **0.0228**

**Answer: (i) 0.6568, (ii) 0.9962, (iii) 0.0228**

## Check yourself

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

1. With variance 2.25 volts, sigma is:
   A. 1.5
   B. 2.25
   C. 4.5
   D. 0.5
   Answer: A. sigma = sqrt(2.25). (Key point: Standard deviation)

2. P(X < 44) uses z =
   A. 2.00
   B. 4.00
   C. 2.67, giving 0.9962
   D. 0.67
   Answer: C. (44 - 40)/1.5 = 2.67. (Key point: Less than 44)

3. P(X > 43) is:
   A. Phi(2.00) = 0.9772
   B. Phi(-2.00)=0.9772
   C. 1 - Phi(2.00) = 0.0228
   D. 0.5 - 0.0228
   Answer: C. Upper tail is 1 minus the table value. (Key point: More than 43)

## Close variants set in other exams

- [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×

## Related in Probability and Statistics

- [Breakdown voltage: normal probabilities (1.5 V)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abv-39-42-le43-ge39) — Probability and Statistics, chapter 3, asked 3×

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- [When terminal voltage is less or more than emf](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aterminal-voltage-less-or-more) — Electric Machines II, chapter 1
- [T/F: more circuits rather than higher voltage](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-circuits-rather-than-voltage) — Transmission and Distribution System Design, chapter 2