A beer company runs trucks along 100 miles of road and has a garage halfway. Each of the trucks breaks down X miles from Hangtown, 100 miles from Dry Gulch. X is a uniform random distribution.
To find a lower bound for the probability, we use the Chebyshev inequality \[ P(|X-\mu | \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2} \\ = P(|X-50 | \geq 10) \leq \frac{\sigma^2}{10^2} \\ \] Since the distribution is uniform, we know the variance \[ \sigma^2 = \frac{(b-a)}{\sqrt{12}} = \frac{100}{\sqrt{12}} = 28.9 \implies \\ P(|X-50| \geq 10) \leq \frac{28.9}{100} = .289 \]
Suppose that in one bad week, 20 trucks break down. Find the lower bound for the probability \(P(A_{20}-50) \leq 10)\) where \(A_20\) is the average of the distances from Hangtown to the break-down point.
First, we know the standard deviation is \[ \sigma = 5.4 \] That means that a breakdown at 20 miles is several standard deviations from the mean at mile 50. \[ k = \frac{-(50-20)}{5.4} = -5.6 \] We can then use the second formation of the Chebyshev equation to solve the problem: \[ P(|X-\mu | \geq k\sigma) \leq \frac{1}{k^2} \implies \\ P(|X-\mu | \geq k\sigma) \leq \frac{1}{-5.6^2} = .032 \] # Problem 2
Calculate the E(X) and the Var of the binomial distribution using the MGF
The MGF is \[ (1-p+pe^t)^n \] We can find the mean by calculating \(M'(0)\) \[ M'(0) = n(1-p+pe^0)^{n-1}(pe^0) \\ = n(1-p+p)^n-1(p) \\ = np \] We can find the variance by evaluating the second derivative at 0 \[ M''(0) = n(1-p+pe^0)^{n-1}(pe^0)+(pe^0)n(n-1)(1-p+pe^0)^{n-2}(pe^0) \\ = n(1)p + pn(n-1)p \\ = np + n^2p^2 \] Then we have to subtract the squared mean from this value \[ \sigma^2 = M''(0) - (M'(0)^2) \\ =np+n^2p^2-np^2-(np)^2 \\ = np - np^2 \\ = np(1-p) \] # Problem 3
Calculate the E(X) and the Var of the exponential distribution using the MGF
The MGF is \[ \frac{1}{1 - \lambda t} \] We can find the mean by calculating \(M'(0)\) \[ M'(0) = -(1-\lambda \cdot 0) ^ -2 * -\lambda \\ = \lambda \] We can find the variance by evaluating the second derivative at 0 \[ M''(0) = \lambda\cdot-2(1-\lambda t)^-3 * -\lambda \\ = 2 \lambda^2 \] Then we have to subtract the squared mean from this value \[ \sigma^2 = M''(0) - (M'(0)^2) \\ = 2 \lambda^2 - \lambda^2 \\ = \lambda^2 \]