The Pilsdorff beer company runs a fleet of trucks along the 100 mile road from Hangtown to Dry Gulch, and maintains a garage halfway in between. Each of the trucks is apt to break down at a point X miles from Hangtown, where X is a random variable uniformly distributed over [0, 100].
For reference: Chebyshev’s Inequality: \(P(|X - \mu|\geq k\sigma)\leq \frac{1}{k^2},k>0\)
Find a lower bound for the probability \(P(|x - 50| \leq 10\)
\(\mu = 50\),\(k\sigma = 10\):
\(\sigma = \frac{100}{\sqrt{{12}}} = 28.87\), \(k = \frac{10}{\sigma}=0.35\)
So: \(P(|x - 50| \leq 10 = \frac{1}{k^2} = \frac{1}{0.122} < 1\)
Suppose that in one bad week, 20 trucks break down. Find a lower bound for the probability \(P(|A_{20}−50|≤10)\), where \(A_20\) is the average of the distances from Hangtown at the time of breakdown.
Again : \(\mu = 50, \epsilon = 10, n = 20\), \(\sigma^2 = \frac{(b -a)^2}{12}=\frac{10000}{12}=833.3\)
So: \(P(|A_{20}−50|≤10) = \frac{833.3}{20(10^2)}=0.416\)
Thus:\(P(|A_{20}−50|≤10)>1-0.416 = 0.58\)