Question

Question: Use Chebyshev's inequality to find an upper bound on the probability that the number of tails that come up when a fair coin is tossed \(n\) times deviates from the mean by more than \(5\sqrt n \).

Short answer

Answer

The upper bound on the probability is \(\frac{1}{{100}}\).

Step-by-step solution

In the problem given  

Number of tails that come up when a fair coin is tossed time deviates from the mean by more than \(5\sqrt n \).

The definition and the formula for the given problem

Chebyshev's inequality: it states that minimum of just \(75\% \) of values must lie within two standard deviations of the mean and \(89\% \) within three standard deviations.

Determining the sum in expanded form

Let \(x\) be the random variable that counts the number of tails when a fair coin is tossed \(n\) times.

Note that \(x\) is the number of successes when on independent Bernoulli traits each with probability of success \(\frac{1}{2}\) are performed.

It follows that, \(E(x) = \frac{n}{2}\) and \({\rm{V}}(x) = \frac{n}{4}\)

Apply Chebyshev's inequality with, \(r = 5\sqrt n \).

\(\begin{aligned}{c}P\left( {1 \times (s) - \frac{n}{2}\mid \ge 5\sqrt n } \right) \le \frac{n}{4}\mid {(5\sqrt n )^2} &= \frac{n}{4} \times \frac{1}{{25n}}\\ &= \frac{1}{{100}}\end{aligned}\)

So, the probability is no more than \(\frac{1}{{100}}\).

The probability that the plant will fill between 9,000 and 11,000 can is \(\frac{{999}}{{1.000}}\).