---
title: "Sample mean income of 10 retailers"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:retailers-income-sample-10"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/retailers-income-sample-10
site: BCT Engineering
---

# Sample mean income of 10 retailers

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2071 Shrawan, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/retailers-income-sample-10
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

The monthly income of a particular group of retailer's follows a normal distribution with mean Rs.21,000.00 and standard deviation of Rs.9,487.00. A random sample of size 10 retailers was taken and the mean income is calculated. Find the probability that this sample lies between Rs.18,000.00 and Rs.27,000.00.

## Short answer

Since the population is normal, sample mean is normal with mean Rs 21,000 and SE=9487/√10≈Rs 3,000. z1=-1.00 and z2=2.00, so P(18,000<x̄<27,000)=0.3413+0.4772=0.8186.

## Answer

Income X ~ N(μ = Rs 21,000, σ = Rs 9,487), n = 10. Since the population is normal, x̄ is normal.

Standard error = σ/√n = 9487/√10 = 9487/3.1623 = Rs 3000 (about Rs 3,000)

- z₁ = (18,000 − 21,000)/3,000 = -1.00
- z₂ = (27,000 − 21,000)/3,000 = 2.00

P(18,000 < x̄ < 27,000) = P(−1 < Z < 2) = P(0 < Z < 1) + P(0 < Z < 2)

= 0.3413 + 0.4772 = **0.8186**

**Answer: the probability is about 0.8186 (81.86%).**

## Check yourself

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

1. With σ = Rs 9,487 and n = 10, the standard error of the mean income is about:
   A. Rs 949
   B. Rs 9,487
   C. Rs 30,000
   D. Rs 3,000
   Answer: D. SE = 9487/√10 = 9487/3.1623 = about Rs 3,000. (Key point: Standard error)

2. Why can we use the normal distribution for x̄ although n = 10?
   A. The population is normal, so x̄ is normal for any n
   B. Because n > 5
   C. Because the standard error is a round number
   D. Because the mean income is large
   Answer: A. A normal parent population gives an exactly normal sample mean for any n. (Key point: Normal population, small n)

3. What are the z values for 18,000 and 27,000?
   A. -0.32 and 0.63
   B. -1 and 3
   C. -3 and 6
   D. -1 and 2
   Answer: D. z1 = (18000 − 21000)/3000 = −1 and z2 = (27000 − 21000)/3000 = 2. (Key point: z-limits)

4. How is P(−1 < Z < 2) obtained from the table?
   A. 0.4772 − 0.3413
   B. 0.3413 + 0.4772 = 0.8186
   C. 0.5 + 0.4772
   D. 2 × 0.4772
   Answer: B. The limits lie on opposite sides of zero, so the two areas are added: 0.3413 + 0.4772 = 0.8186. (Key point: Area between z limits)

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