---
title: "CLT: bulb mean life between 1100 and 1300"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-bulb-1200-hours-between-1100"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-bulb-1200-hours-between-1100
site: BCT Engineering
---

# CLT: bulb mean life between 1100 and 1300

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2070 Chaitra, 2+4 marks
- Video: [The Central Limit Theorem, Clearly Explained!!!](https://www.youtube.com/watch?v=YAlJCEDH2uY) — StatQuest with Josh Starmer
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-bulb-1200-hours-between-1100
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

The lifetime of a certain brand of an electric bulb may be considered a random variable with mean 1200 hours and standard deviation 150 hours. Using central limit theorem, find the probability that the sample mean of the lifetime with a sample of size 36, is between 1100 hours and 1300 hours.

## Short answer

By CLT, x̄ ~ N(1200, 25²) with SE = 150/√36 = 25. z-scores are -4 and 4, so P(1100 < x̄ < 1300) = P(-4 < Z < 4) ≈ 0.99994.

## Answer

**Central Limit Theorem (CLT):** if X₁, X₂, …, Xₙ is a random sample of size *n* from any population with mean μ and finite variance σ², then for large *n* (n ≥ 30) the sampling distribution of the sample mean x̄ is approximately normal with mean μ and standard deviation σ/√n, whatever the shape of the population:

x̄ ~ N(μ, σ²/n), and z = (x̄ − μ)/(σ/√n) ~ N(0, 1).

If the population itself is normal, x̄ is exactly normal for every n.

#### Numerical
μ = 1200 h, σ = 150 h, n = 36 (large enough for CLT).

SE = σ/√n = 150/6 = 25 hours, so x̄ ~ N(1200, 25²)

- z₁ = (1100 − 1200)/25 = -4.00
- z₂ = (1300 − 1200)/25 = 4.00

P(1100 < x̄ < 1300) = P(−4 < Z < 4) = 2 × P(0 < Z < 4) ≈ 2 × 0.49997 = **0.99994**

**Answer: the probability is about 0.99994, almost certain.**

## Check yourself

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

1. With σ = 150 hours and n = 36, the standard error of x̄ is:
   A. 25 hours
   B. 4.17 hours
   C. 150 hours
   D. 6 hours
   Answer: A. SE = 150/√36 = 150/6 = 25 hours. (Key point: Standard error)

2. What are the z values for x̄ = 1100 and 1300?
   A. -0.67 and 0.67
   B. -4 and 4
   C. -100 and 100
   D. -2 and 2
   Answer: B. z = (1100 − 1200)/25 = −4 and (1300 − 1200)/25 = 4. (Key point: z limits)

3. The population shape is not stated. Why can the normal distribution be used for x̄?
   A. Because σ is known
   B. Because the sample size 36 is large (n ≥ 30), so CLT gives approximate normality
   C. Because the mean is 1200
   D. Because the probability is symmetric
   Answer: B. For n ≥ 30 the CLT makes x̄ approximately normal regardless of the population shape. (Key point: Why CLT applies)

4. P(−4 < Z < 4) is about:
   A. 0.5
   B. 0.68
   C. 0.99994
   D. 0.0001
   Answer: C. 2 × 0.49997 = 0.99994, so the sample mean almost surely lies between 1100 and 1300. (Key point: Final probability)

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