---
title: "CLT: sample mean between 75 and 78"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-sample-100-mean-76-variance"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-sample-100-mean-76-variance
site: BCT Engineering
---

# CLT: sample mean between 75 and 78

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2068 Chaitra, 6 marks
- Video: [Central Limit Theorem (CLT) | Statistical Inference | VTU Exam Problem | Part 4](https://www.youtube.com/watch?v=EVlUlUnWbBk) — Mathematics Tutor
- 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-76-variance
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

State the central limit theorem. A random sample of size 100 is taken from an infinite population having the mean 76 and variance 256. What is the probability that the sample mean will be between 75 and 78?

## Short answer

By CLT, x̄ ~ N(76, 1.6²) with SE = 16/√100 = 1.6. z-scores: (75-76)/1.6 = -0.625, (78-76)/1.6 = 1.25. P(75 < x̄ < 78) = Φ(1.25) - Φ(-0.625) = 0.8944 - 0.2660 = 0.6284.

## Answer

#### Central limit theorem

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* (usually n ≥ 30) the sampling distribution of the sample mean x̄ is approximately normal with mean μ and variance σ²/n, whatever the shape of the parent population. That is,

Z = (x̄ − μ)/(σ/√n) → N(0, 1) as n → ∞.

#### Numerical

Given: n = 100, μ = 76, σ² = 256, so σ = 16.

Since n is large, x̄ ~ N(μ, σ²/n):

- Mean of x̄ = 76
- Standard error σ/√n = 16/√100 = 16/10 = 1.6

Standardise:

- x̄ = 75: Z₁ = (75 − 76)/1.6 = -0.625
- x̄ = 78: Z₂ = (78 − 76)/1.6 = 1.250

P(75 < x̄ < 78) = P(-0.625 < Z < 1.250) = Φ(1.250) − Φ(-0.625)
= 0.8944 − 0.2660 = 0.6284

**Answer: P(75 < x̄ < 78) ≈ 0.6284 (about 62.8%).**

## Check yourself

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

1. Variance is 256 and n = 100. The standard error of x̄ is:
   A. 16
   B. 2.56
   C. 1.6
   D. 0.16
   Answer: C. σ = 16, so SE = 16/√100 = 1.6. (Key point: Standard error)

2. Why is x̄ approximately normal here?
   A. Because the variance is 256
   B. Because the mean is 76
   C. Because the population is infinite
   D. Because n = 100 is large, so the CLT applies
   Answer: D. With n ≥ 30 the CLT gives x̄ ~ N(μ, σ²/n) whatever the population shape. (Key point: Why normal)

3. The z values for x̄ = 75 and 78 are:
   A. -0.625 and 1.25
   B. -1 and 2
   C. -0.0625 and 0.125
   D. -1.6 and 3.2
   Answer: A. (75 − 76)/1.6 = −0.625 and (78 − 76)/1.6 = 1.25. (Key point: z-limits)

4. P(−0.625 < Z < 1.25) is found as:
   A. Φ(1.25) + Φ(−0.625)
   B. Φ(1.25) − Φ(−0.625) = 0.8944 − 0.2660 = 0.6284
   C. Φ(1.25) × Φ(0.625)
   D. 1 − Φ(1.25)
   Answer: B. The probability between two limits is the difference of the cumulative areas, 0.6284. (Key point: Probability between limits)

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