Question 1178271
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).