---
title: "CLT: sample mean between 67 and 83"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-sample-100-mean-75-variance"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-sample-100-mean-75-variance
site: BCT Engineering
---

# CLT: sample mean between 67 and 83

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2074 Chaitra, 6 marks
- Video: [Applying the central limit theorem to find probability example 3](https://www.youtube.com/watch?v=MBU6Mc5Bf9Y) — Dane McGuckian (STATSprofessor)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-sample-100-mean-75-variance
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

State central limit theorem. A random sample of size 100 is taken from an infinite population with mean 75 and variance 256. Assert the chances of sample mean between 67 and 83.

## Short answer

The central limit theorem states that for large samples the sample mean is approximately normal with mean μ and variance σ²/n. Here μ=75, σ=16, SE=1.6; z values are ±5, so P(67<x̄<83) is about 0.9999994, practically 1.

## 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
Given μ = 75, σ² = 256, so σ = 16, and n = 100 (large, so CLT applies).

Standard error = σ/√n = 16/√100 = 1.6

x̄ ~ N(75, 1.6²)

z₁ = (67 − 75)/1.6 = -5.00  and  z₂ = (83 − 75)/1.6 = 5.00

P(67 < x̄ < 83) = P(−5 < Z < 5) = 2 × P(0 < Z < 5) ≈ 2 × 0.4999997 = 0.9999994

**Answer: the probability that the sample mean lies between 67 and 83 is about 0.9999994, i.e. practically 1 (certain).**

## Check yourself

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

1. With variance 256 and n = 100, SE of x̄ is:
   A. 16
   B. 2.56
   C. 1.6
   D. 25.6
   Answer: C. σ = 16, so SE = 16/sqrt(100) = 1.6. (Key point: Standard error)

2. The z-values for 67 and 83 (mean 75) are:
   A. −5 and 5
   B. −0.5 and 0.5
   C. −8 and 8
   D. −1.6 and 1.6
   Answer: A. z = ±8/1.6 = ±5. (Key point: z-values)

3. Why can the normal table be used here?
   A. n = 100 is large, so the CLT applies
   B. The population is known to be normal
   C. The variance is large
   D. The mean is 75
   Answer: A. For large n, x̄ is approximately normal regardless of the population shape. (Key point: Why normal applies)

4. P(67 < x̄ < 83) is about:
   A. 0.5
   B. 0.68
   C. 0.95
   D. practically 1
   Answer: D. The limits are ±5 standard errors, so almost all the area (0.999999) lies inside. (Key point: Result)

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