---
title: "CLT: bulbs average life below 775 h"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-bulb-sample-mean-below-775"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-bulb-sample-mean-below-775
site: BCT Engineering
---

# CLT: bulbs average life below 775 h

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2071 Chaitra, 5 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-sample-mean-below-775
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

State central limit theorem. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with mean equal to 800 hours and standard deviation of 4 hours. Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours. [Printed as "12775" hours in the scan.]

## Short answer

For a normal population, sample mean is normal with mean 800 h and SE=4/√16=1 h. z=(775-800)/1=-25, so P(x̄<775) is practically zero.

## 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 = 800 hours, σ = 4 hours, n = 16 (the population is normal, so x̄ is normal).

SE = σ/√n = 4/√16 = 1 hour

z = (775 − 800)/1 = -25

P(x̄ < 775) = P(Z < −25) ≈ 0

**Answer: the probability is practically zero (less than 0.0001).**

*Note:* the scanned question shows "12775" hours, read here as 775 hours. If σ were 40 hours (as in the similar question), SE = 10, z = −2.50 and P = 0.0062.

## Check yourself

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

1. For σ = 4 hours and n = 16 bulbs, the standard error of the sample mean is:
   A. 4 hours
   B. 0.25 hours
   C. 16 hours
   D. 1 hour
   Answer: D. SE = σ/√n = 4/√16 = 4/4 = 1 hour. (Key point: Standard error for n = 16)

2. What is z for x̄ = 775 when μ = 800 and SE = 1?
   A. -25
   B. -6.25
   C. 25 with area 0.5 on the right
   D. -1
   Answer: A. z = (775 − 800)/1 = −25. (Key point: z-value for sample mean)

3. Here n = 16 is below 30, yet x̄ can still be treated as normal. Why?
   A. Because the standard error is small
   B. Because any sample mean is normal
   C. Because the CLT works for every n
   D. Because the population itself is normal, so x̄ is exactly normal for any n
   Answer: D. If the parent population is normal, the sample mean is exactly normal for every n. (Key point: Normality for small n)

4. What is P(x̄ < 775)?
   A. About 0.5
   B. Practically zero (less than 0.0001)
   C. About 0.0062
   D. About 0.9938
   Answer: B. z = −25 lies far in the lower tail, so the probability is practically zero. (Key point: Result of the probability)

## Close variants set in other exams

- [CLT: average life of 16 bulbs](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-bulbs-16-850-750-900) — Probability and Statistics, chapter 4, asked 2×

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