---
title: "CLT: average resistance of 25 resistors"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-resistors-mean-below-95"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-resistors-mean-below-95
site: BCT Engineering
---

# CLT: average resistance of 25 resistors

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2072 Kartik, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-resistors-mean-below-95
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

Define Central Limit Theorem. An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of 25 resistors will have an average resistance less than 95 ohms.

## Short answer

By CLT, sample mean is normal with mean 100 Ω and SE=10/√25=2 Ω. z=(95-100)/2=-2.50, so P(x̄<95)=0.0062.

## 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 = 100 Ω, σ = 10 Ω, n = 25, population normal.

SE = σ/√n = 10/5 = 2 Ω

z = (95 − 100)/2 = -2.50

P(x̄ < 95) = P(Z < −2.50) = 0.5 − 0.4938 = **0.0062**

**Answer: the probability that the average resistance is below 95 Ω is 0.0062 (about 0.62%).**

## Check yourself

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

1. For σ = 10 Ω and n = 25 resistors, the standard error of x̄ is:
   A. 2 Ω
   B. 10 Ω
   C. 0.4 Ω
   D. 5 Ω
   Answer: A. SE = 10/√25 = 10/5 = 2 Ω. (Key point: Standard error)

2. What is the z value for an average resistance of 95 Ω?
   A. -0.5
   B. -5
   C. -2.50
   D. 2.50
   Answer: C. z = (95 − 100)/2 = −2.50. (Key point: z-score)

3. How is P(Z < −2.50) found from a table that gives area from 0 to z (0.4938)?
   A. 0.4938
   B. 0.5 + 0.4938
   C. 1 − 0.4938
   D. 0.5 − 0.4938 = 0.0062
   Answer: D. The area to the left of −2.50 is the tail beyond z, so 0.5 − 0.4938 = 0.0062. (Key point: Tail probability)

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