SOLUTION: How large should the size of a random sample be so that we can be 90% certain that the sample mean X will not deviate from the true mean by more than σ/2? Thank you :)

Question 1178061: How large should the size of a random sample be so that we can be 90% certain that the sample mean X will not deviate from the true mean by more than σ/2?
Thank you :)
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
Let's solve this problem using Chebyshev's Inequality.
**Understanding the Problem**
* **Confidence Level:** 90% (0.90)
* **Deviation:** The sample mean (X̄) should not deviate from the true mean (μ) by more than σ/2.
* **Goal:** Find the sample size (n).
**Chebyshev's Inequality**
Chebyshev's Inequality states:
* P(|X̄ - μ| ≥ kσ_X̄) ≤ 1/k²
* Or, P(|X̄ - μ| < kσ_X̄) ≥ 1 - 1/k²
Where:
* X̄ is the sample mean
* μ is the true mean
* σ_X̄ is the standard deviation of the sample mean (σ/√n)
* k is a positive constant
**Applying Chebyshev's Inequality**
1. **Set Up the Inequality:**
* We want P(|X̄ - μ| ≤ σ/2) ≥ 0.90.
* This is equivalent to P(|X̄ - μ| < σ/2) ≥ 0.90.
* We know that σ_X̄ = σ/√n.
* So, we have P(|X̄ - μ| < kσ/√n) ≥ 0.90.
2. **Find k:**
* We are given that |X̄ - μ| ≤ σ/2.
* Comparing this with |X̄ - μ| < kσ/√n, we have:
* kσ/√n = σ/2
* k/√n = 1/2
* k = √n / 2
3. **Use Chebyshev's Inequality:**
* 1 - 1/k² ≥ 0.90
* 1 - 0.90 ≥ 1/k²
* 0.10 ≥ 1/k²
* k² ≥ 1/0.10 = 10
* k ≥ √10
4. **Substitute k:**
* √n / 2 ≥ √10
* √n ≥ 2√10
* n ≥ (2√10)²
* n ≥ 4 * 10
* n ≥ 40
**Conclusion**
The sample size should be at least 40 so that we can be 90% certain that the sample mean X̄ will not deviate from the true mean μ by more than σ/2.

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