pg.322-323 8.2.11
11.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].
(a) Find a lower bound for the probability \(P(|X-50|\leq10)\).
Chebyshev Inequality: \(P(|X-\mu|\geq k\sigma)\leq\frac{1}{k^2}\)
In our case, \(\mu=50\) and \(k\sigma=10\)
In a uniform distribution, \(\sigma=\frac{(b-a)}{\sqrt(12)}\), so \(\sigma=\frac{100}{\sqrt(12)}=28.87\). \(k=\frac{10}{28.87}=0.35\)
\(P(|X-50|\geq 10)\leq\frac{1}{0.35^2}\)
\(P(|X-50|\geq 10)\leq\frac{1}{0.1225}\)
So, it follows that \(P(|X-50| < 10)\geq 1-\frac{1}{0.1225}=0.8775\)
(b) Suppose that in one bad week, 20 trucks break down. Find a lower bound for the probability \(P(|A_{20}-50|\leq10)\), where \(A_{20}\) is the average of the distances from Hangtown at the time of breakdown.
In this problem, we are considering average distances (\(A_{20}\)), so Chebyshev’s Inequality for any \(\epsilon>0\) will be \(P(|\frac{S_{n}}n-\mu|\geq\epsilon)\leq\frac{\sigma^2}{n\epsilon^2}\) where \(\mu=50\), \(\epsilon=10\), \(n=20\) and \(\sigma^2=\frac{(b-a)^2}{12}=833.33\).
\(P(|A_{20}-50|\geq10)\leq\frac{833.33}{(20)10^2}\)
\(P(|A_{20}-50|\geq10)\leq 0.4167\)
So, the lower bound would be \(P(|A_{20}-50|<10)\geq 1-0.4167 = 0.5833\)