---
title: "CLT: average impurity of 50 batches"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-impurity-50-batches"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-impurity-50-batches
site: BCT Engineering
---

# CLT: average impurity of 50 batches

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2074 Asoj, 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-impurity-50-batches
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

Define Central Limit Theorem. The amount of impurity in a batch of a certain chemical product is a random variable with mean value 4.0 gm and standard deviation 1.5 gm. If 50 batches are independently prepared, what is the probability that the sample average amount of impurity is between 3.5 and 3.8 gm?

## Short answer

For large samples, the sample mean is approximately normal with mean μ and standard error σ/√n. Here SE=1.5/√50=0.2121 g; z1=-2.36, z2=-0.94, so P(3.5<x̄<3.8)=0.1736-0.0091=0.1645.

## 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 = 4.0 g, σ = 1.5 g, n = 50 (large, so CLT applies).

SE = σ/√n = 1.5/√50 = 1.5/7.0711 = 0.2121

x̄ ~ N(4.0, 0.2121²)

- z₁ = (3.5 − 4.0)/0.2121 = -2.36
- z₂ = (3.8 − 4.0)/0.2121 = -0.94

P(3.5 < x̄ < 3.8) = P(−2.36 < Z < −0.94) = P(Z < −0.94) − P(Z < −2.36)

= 0.1736 − 0.0091 = **0.1645**

**Answer: the probability is about 0.1645.**

## Check yourself

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

1. For σ = 1.5 and n = 50, SE is about:
   A. 0.2121
   B. 0.03
   C. 1.5
   D. 0.75
   Answer: A. 1.5/sqrt(50) = 1.5/7.0711 = 0.2121. (Key point: Standard error)

2. The z-values for 3.5 and 3.8 (mean 4.0) are about:
   A. −2.36 and −0.94
   B. −0.5 and −0.2
   C. 2.36 and 0.94
   D. −2.36 and 0.94
   Answer: A. z = (3.5 − 4)/0.2121 = −2.36 and (3.8 − 4)/0.2121 = −0.94. (Key point: z-values)

3. P(−2.36 < Z < −0.94) is computed as:
   A. P(Z < −0.94) − P(Z < −2.36) = 0.1645
   B. 0.4909 + 0.3264
   C. 0.5 − 0.4909
   D. P(Z < −2.36) − P(Z < −0.94)
   Answer: A. Subtract the smaller left-tail area from the larger: 0.1736 − 0.0091. (Key point: Area between two negative z)

## Related in Probability and Statistics

- [CLT: bulbs average life below 775 h](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-bulb-sample-mean-below-775) — Probability and Statistics, chapter 4, asked 2×
- [CLT: average resistance of 25 resistors](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-resistors-mean-below-95) — Probability and Statistics, chapter 4
- [CLT: battery cell average life](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abattery-cells-clt-more-15-less) — Probability and Statistics, chapter 4
- [CLT: bulb mean life between 1100 and 1300](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-bulb-1200-hours-between-1100) — Probability and Statistics, chapter 4
- [CLT: sample mean between 67 and 83](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-sample-100-mean-75-variance) — Probability and Statistics, chapter 4, asked 2×
- [CLT: sample mean between 75 and 78](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-sample-100-mean-76-variance) — Probability and Statistics, chapter 4, asked 2×

## Related in other subjects

- [Moments of random variable; NEB of RC-LPF; output noise power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amoments-neb-rc-lpf) — Communication Systems, chapter 9
- [Standard deviation and mean of array](https://www.bctengineer.com/question.md?key=semester_1%2Fcomputer_programming%3Astandard-deviation-pseudocode) — Computer Programming, chapter 7
- [Random process; LTI output is WSS if input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arandom-process-lti-wssp) — Communication Systems, chapter 9
- [HR trickling filter, 2.5 MLD; change if standard](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-hr-2p5mld-std-compare) — Sanitary Engineering, chapter 7