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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

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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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    Why is p = 0.5? (Just trying to understand the question a little better as it won't be identical tomorrow!)2012-08-14
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    @Panda Because you said you are using a fair coin. Probability of success here (named $p$) is probability of tossing a head, which must equal the probability of tossing a tail, since the coin is fair. There are only 2 outcomes of a toss, with equal probabilities, hence each must be $1/2$.2012-08-14
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    I've gotten terrible at this stuff. Would've seen that stuff straight away before! Thanks so much! Gonna be a long night :'(2012-08-14
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    @gt6989b: There may be a few typos in the solution.2012-08-14
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    What is the difference between an upper bound and a lower bound? Are they the same? I thought the OP wanted a **lower** bound on $P\{60 < X < 80\}$ but you have provided an **upper** bound instead since you get $$P\{60 < X < 80\} < \left(\frac{4}{49}\right).$$2012-08-15
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    @DilipSarwate Thanks for the note. Will need to fix the answer - but in short, an upper bound of probability $p$ is a lower bound on $1-p$...2012-08-15
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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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    How do you know how to recognise questions as Binomial or Normal etc?2012-08-14
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    When you learn Bernoulli trials, you learn why their distributions are binomial. How did I identify it as a Bernoulli experiement then? I'm using the mathematical principle: "If it looks like a duck and quacks like a duck, it's a duck." That is to say *it satisfies the definition of what a Bernoulli trial is*. I think you need to reread your text section on Bernoulli trials, as well as their explanation of how the binomial distribution comes out of it.2012-08-14
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    I suppose I should add one thing to that: you probably learned that the normal distribution can approximate the binomial distribution, so you might me confused as to when you would want to use the normal distribution. It is advantageous to use the normal approximation if the number of trials is really high, or else the number of individual events you need to compute is large.2012-08-14
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    @Panda Binomial distribution is a sum of Bernoulli trials. Normal is typically modeling some continuous quantity, e.g., error of some sort. For example, amount of soup automatically dispensed into a container would vary somewhat, and it is typical to model as Normal.2012-08-14
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    @rschwieb I am a bit rusty on this stuff it seems. It's been a year or so since I learned that stuff :/ For Chebychev's Inequality what are the usual distributions to come up in it would you say? Similar difficulty to this - I want to be able to recognise them tomorrow.2012-08-14
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    Just saw the two comments above! Thanks2012-08-14
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    @Panda The beauty of Chebyshev's Inequality is that it works for all distributions :). Anything where you can compute expected value and variance will do.2012-08-14
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    @gt6989b Okay so I have some brushing up to do! Thanks again! :)2012-08-14
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    @rschwieb : If the problem explicitly said you should get a bound by using Chebyshev's inequality, then your answer is wrong. I.e., although it's correct, it doesn't address the question actually asked.2012-08-14
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    @gt6989b : It works for all distributions that have a finite variance. Not for others.2012-08-14
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    @MichaelHardy I think you definitely misread it. You have to know it's the binomial distribution so that you can use $\mu$ and $\sigma$ in Chebyshev's formula. I said "use that data to fill in what you wrote" [OP wrote Chebyshev's bound out]. Then the OP applies complementary probability and is done. Very direct. The rest of the comments are general advice addressed to the OP's questions in the comments.2012-08-14
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    OK, I see what you're saying.2012-08-14