---
title: "Normal: storage battery life"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:battery-life-normal-2-3-2"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/battery-life-normal-2-3-2
site: BCT Engineering
---

# Normal: storage battery life

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2068 Jestha (old course), 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/battery-life-normal-2-3-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will (i) last less than 2.3 years, and (ii) will last at most 2.5 years.

## Short answer

For X~N(3,0.5²): (i) z=(2.3−3)/0.5=−1.4, P(X<2.3)=0.0808. (ii) z=(2.5−3)/0.5=−1.0, P(X≤2.5)=0.1587.

## Answer

Battery life X ~ N(μ = 3.0 years, σ = 0.5 year).

#### (i) Less than 2.3 years
z = (2.3 − 3.0)/0.5 = -1.40

P(X < 2.3) = P(Z < −1.40) = 0.5 − P(0 < Z < 1.40) = 0.5 − 0.4192 = **0.0808**

#### (ii) At most 2.5 years
z = (2.5 − 3.0)/0.5 = -1.00

P(X ≤ 2.5) = P(Z ≤ −1.00) = 0.5 − 0.3413 = **0.1587**

**Answer: (i) 0.0808 (8.08%)  (ii) 0.1587 (15.87%)**

## Check yourself

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

1. For N(3.0, 0.5), z at 2.3 years is:
   A. −1.40
   B. −0.70
   C. +1.40
   D. −0.14
   Answer: A. z = (2.3 − 3.0)/0.5 = −1.40. (Key point: z for 2.3 years)

2. P(Z < −1.40) with table area 0.4192 is:
   A. 0.4192
   B. 0.5 + 0.4192
   C. 0.5 − 0.4192 = 0.0808
   D. 1 − 0.0808
   Answer: C. The left tail is 0.5 minus the area from 0 to 1.40. (Key point: Left tail)

3. P(X ≤ 2.5) equals:
   A. 0.3413
   B. 0.8413
   C. 0.1587
   D. 0.0808
   Answer: C. z = −1.00, so P = 0.5 − 0.3413 = 0.1587. (Key point: At most 2.5 years)

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