Question 1178060: Suppose that the number of cars arriving at a busy intersection in a large city has a Poisson distribution with mean 120. Determine a lower bound for the probability that the numberof cars arriving in a given 20-minute period will be between 100 and 140 using Chebyshev’s inequality
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Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step using Chebyshev's Inequality.
**Understanding the Problem**
* **Poisson Distribution:** The number of cars arriving follows a Poisson distribution.
* **Mean (λ):** 120 cars per hour.
* **Time Period:** 20 minutes (1/3 of an hour).
* **Goal:** Find a lower bound for P(100 ≤ X ≤ 140), where X is the number of cars arriving in 20 minutes.
**1. Adjust the Mean for the Time Period**
* Since the mean is 120 cars per hour, the mean for a 20-minute period is:
* μ = 120 * (20/60) = 120 * (1/3) = 40 cars
**2. Properties of Poisson Distribution**
* For a Poisson distribution, the mean (μ) and variance (σ²) are equal:
* μ = 40
* σ² = 40
* σ = √40 ≈ 6.3246
**3. Apply Chebyshev's Inequality**
Chebyshev's Inequality states:
* P(|X - μ| ≥ kσ) ≤ 1/k²
* Or, P(|X - μ| < kσ) ≥ 1 - 1/k²
We want to find a lower bound for P(100 ≤ X ≤ 140).
Let's rewrite this as:
* |X - μ| < kσ
* |X - 40| < k√40
We need to find the range of X:
* 140 - 40 = 100
* 40 - 100 = -60. Thus |40 - 100| = 60.
We want to find k such that:
* |X - 40| < 60
Now, find k:
* k√40 = 60
* k = 60 / √40 = 60 / (2√10) = 30 / √10 = 3√10 ≈ 9.4868
Now, apply Chebyshev's Inequality:
* P(|X - 40| < 60) ≥ 1 - 1/k²
* P(100 ≤ X ≤ 140) ≥ 1 - 1/(3√10)²
* P(100 ≤ X ≤ 140) ≥ 1 - 1/90
* P(100 ≤ X ≤ 140) ≥ 89/90 ≈ 0.9889
**Therefore, a lower bound for the probability that the number of cars arriving in a given 20-minute period will be between 100 and 140 is 89/90 or approximately 0.9889.**
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