---
title: "CLT: sample of 16, P(x̄<160)"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:clt-sample-16-mean-150-variance"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-sample-16-mean-150-variance
site: BCT Engineering
---

# CLT: sample of 16, P(x̄<160)

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2070 Asar, 2+4 marks
- Video: [Find the Probability of Sample Means](https://www.youtube.com/watch?v=JtaKNSrCgXU) — Math and Stats Help
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-sample-16-mean-150-variance
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

Define Central Limit Theorem. In a sample of 16 observations from a normal distribution with mean of 150 and a variance of 256, what is (a) P(x̄ < 160) (b) P(x̄ > 142)

## Short answer

CLT: for large n, sample mean is approximately N(μ, σ²/n). With normal population, x̄ ~ N(150, 4²) exactly; (a) z=2.5 gives P=0.9938, (b) z=-2 gives P=0.9772.

## 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
Population normal with μ = 150, σ² = 256 so σ = 16; n = 16.

SE = σ/√n = 16/4 = 4, so x̄ ~ N(150, 4²) (exactly normal since the population is normal).

#### (a) P(x̄ < 160)
z = (160 − 150)/4 = 2.50

P(x̄ < 160) = P(Z < 2.50) = 0.5 + 0.4938 = **0.9938**

#### (b) P(x̄ > 142)
z = (142 − 150)/4 = -2.00

P(x̄ > 142) = P(Z > −2.00) = 0.5 + 0.4772 = **0.9772**

**Answer: (a) 0.9938  (b) 0.9772**

## Check yourself

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

1. The variance is 256 and n = 16. What is the standard error of x̄?
   A. 16
   B. 1
   C. 4
   D. 256/16 = 16
   Answer: C. σ = √256 = 16, so SE = 16/√16 = 4. (Key point: Standard deviation from variance)

2. The z value for x̄ = 160 is:
   A. 10
   B. 0.625
   C. 2.50
   D. -2.50
   Answer: C. z = (160 − 150)/4 = 2.50, giving P = 0.5 + 0.4938 = 0.9938. (Key point: z for P(x̄ < 160))

3. For P(x̄ > 142), z = −2. The probability is:
   A. 0.0228
   B. 0.4772
   C. 0.9772
   D. 0.9544
   Answer: C. P(Z > −2) = 0.5 + 0.4772 = 0.9772. (Key point: Upper-tail probability)

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