# Differences

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 math-340:m340-f15-hw:hw-28 [2015/11/12 11:54]dcs created math-340:m340-f15-hw:hw-28 [2015/11/15 06:57] (current)dcs 2015/11/15 06:57 dcs 2015/11/12 11:54 dcs created 2015/11/15 06:57 dcs 2015/11/12 11:54 dcs created Line 12: Line 12: - Let $X$ be binomial with parameters $n = 4$ and $p = \frac{1}{2}$. Use Chebyshev'​s inequality to find an upper bound $P(|X - 2| \ge 2)$ and compare with the exact probability. - Let $X$ be binomial with parameters $n = 4$ and $p = \frac{1}{2}$. Use Chebyshev'​s inequality to find an upper bound $P(|X - 2| \ge 2)$ and compare with the exact probability. - - If $Z$ is standard normal, find an upper bound for $P(|Z| \ge 4)$ using Markoff's inequality, Chebyshev'​s inequality, and Chernoff'​s inequality. + - If $Z$ is standard normal, find an upper bound for $P(|Z| \ge 4)$ using Markov's inequality, Chebyshev'​s inequality, and Chernoff'​s inequality. + + Answers: + + - By Chebyshev'​s inequality, $P(|X - 2| \ge 2) \le \frac{1}{4}$. Exactly, $P(|X - 2| \ge 2) = \frac{1}{8}$. + - Markov'​s inequality: $P(|Z| \ge 4) \le \frac{1}{4}\sqrt{\frac{2}{\pi}} = 0.1995$; Chebysev'​s inequality: $P(|Z| \ge 4) \le \frac{1}{16} = 0.06250$; Chernoff'​s inequality: $P(|Z| \ge 4) \le 2e^{-8} = 0.006709$

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