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Let $X$ be the number of heads one would obtain in $140$ flips of a fair coin.

Use Chebychev's Inequality to find a lower bound on the probability $P(60 < X < 80)$.

Okay so Chebychev's Inequality is $P(|X - E(X)| > kσ) \le 1/k^2$ for $ k > 0$, where $σ^2$ is the variance of $X$.

I'm not sure how to fill this in or anything. My probabilty test is tomorrow so help is much appreciated! Descriptive answers would be awesome.

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    Which is the $E(X)$ and $\sigma^2$ in your case? Did you compute them?2012-08-14

3 Answers 3

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Note that $X$ is binomial with $n=140, p = 0.5$ so $\mathbb{E}[X] = np = 70$, $\sigma^2 = Var(X) = np(1-p) = 35,$ giving $\sigma = \sqrt{35} $ .

Now,

$\begin{split} \mathbb{P}[60 < X < 80] &= \mathbb{P}[-10 < X - \mathbb{E}[X] < 10] \\ &= \mathbb{P}[|X - \mathbb{E}[X]| < 10] \\ &= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10}{\sqrt{35}} \sigma]\\ &= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10\sigma}{\sqrt{35}}] \end{split} $

By Chebyshev's Inequality, $\mathbb{P}[X \not \in (60,80)] = \mathbb{P}[|X - \mathbb{E}[X]| > \frac{10\sigma}{\sqrt{35}}] \leq (\frac{\sqrt{35}}{10})^2 = 35/100 = 7/20$.

Hence, $\mathbb{P}[60 < X < 80] = 1 - \mathbb{P}[X \not \in (60,80)] \geq 1-7/20 = 13/20. $

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    Your answer, though it has been accepted by @Panda, is **still** incorrect since you are confusing the value of $\sigma^2$, the _variance_, with the value of $\sigma$, the _standard deviation._ In particular, it is not $\sigma$ that equals $35$ but $\sigma^2$. See André Nicolas's answer for the correct derivation.2012-08-16
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Before the solution, a minor comment. The Chebyshev Inequality is not quite quoted correctly. It should be $\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.\tag{$1$}$ For continuous distributions there is no need to distinguish between $\le$ and $\lt$. Here we are working with a discrete distribution.

A standard calculation shows that in our case $\mu=np=70$ and $\sigma^2=np(1-p)=35$. We want a lower bound on $\Pr(60\lt X\lt 80)$. The complementary event is $|X-70|\ge 10$. We first find an upper bound for $\Pr(|X-70|\ge 10)$.

Compare with Inequality $(1)$ quoted above. In our case we have $k\sigma=10$, and therefore $k=\frac{10}{\sigma}, \quad\text{so}\quad\frac{1}{k^2}=\frac{\sigma^2}{100}=\frac{35}{100}.$

It follows that $\Pr(|X-70|\ge 10)$ is $\le \frac{35}{100}$. Thus $\Pr(60\lt X\lt 80)\ge 1-\frac{35}{100}=\frac{65}{100}.$ That is the lower bound given by the Chebyshev Inequality.

Remark: It is not a very good lower bound. You might want to use software such as the free-to-use Wolfram Alpha to calculate the exact probability. It's not Chebyshev's fault. An inequality that works for every distribution that has a mean and variance, including some pretty weird ones, cannot be expected to compete against estimates based on more information.

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Hint: You should recognize this experiment as a series of Bernoulli trials, and so its probability distribution is given by the binomial distribution! Compute the mean and variance and fill in the blanks and you are basically done!

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    OK, I see what you're saying.2012-08-14