Question 1178271: Suppose that the number of cars arriving in 1 hour at a busy intersection is a Poisson probability distribution with λ = 100. Find, using Chebyshev’s inequality, a lower bound for the probability that the number of cars arriving at the intersection in 1 hour is between 70 and 130.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem using Chebyshev's Inequality.
**Understanding the Problem**
* **Poisson Distribution:** The number of cars arriving in 1 hour follows a Poisson distribution with λ = 100.
* **Mean and Variance:** For a Poisson distribution, the mean (μ) and variance (σ²) are both equal to λ. Therefore, μ = 100 and σ² = 100, so σ = √100 = 10.
* **Interval:** We want to find a lower bound for P(70 ≤ X ≤ 130).
**Chebyshev's Inequality**
Chebyshev's Inequality states that for any random variable X with mean μ and standard deviation σ, and for any k > 0:
* P(|X - μ| ≥ kσ) ≤ 1/k²
We can also write it as:
* P(|X - μ| < kσ) ≥ 1 - 1/k²
**Applying Chebyshev's Inequality**
1. **Find k:**
* We want to find P(70 ≤ X ≤ 130), which is equivalent to P(|X - 100| ≤ 30).
* We need to find k such that kσ = 30.
* Since σ = 10, we have k * 10 = 30, so k = 3.
2. **Apply the Inequality:**
* P(|X - 100| < 30) ≥ 1 - 1/k²
* P(|X - 100| < 30) ≥ 1 - 1/3²
* P(|X - 100| < 30) ≥ 1 - 1/9
* P(|X - 100| < 30) ≥ 8/9
3. **Interpret the Result:**
* P(70 ≤ X ≤ 130) ≥ 8/9
* 8/9 ≈ 0.8889
**Conclusion**
Using Chebyshev's Inequality, a lower bound for the probability that the number of cars arriving at the intersection in 1 hour is between 70 and 130 is 8/9 (approximately 0.8889).
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