Discussion 9

Exercise 4

Let S be the number of heads in 1,000,000 tosses of a fair coin. Use (a) Chebyshev’s inequality, and (b) the Central Limit Theorem, to estimate the probability that S lies between 499,500 and 500,500. Use the same two methods to estimate the probability that S lies between 499,000 and 501,000, and the probability that S lies between 498,500 and 501,500.

Let \(n\) be the number of coin tosses (in this case, \(n = 1,000,000\)). \ Let \(p\) be the probability of getting a head on a single toss (since the coin is fair, \(p = 0.5\)). \ Let \(\mu\) be the mean of the binomial distribution, which is \(n \times p\). \ Let \(\sigma^2\) be the variance of the binomial distribution, which is \(n \times p \times (1-p)\). \ Let \(\sigma\) be the standard deviation of the binomial distribution, which is \(\sqrt{n \times p \times (1-p)}\).

We want to estimate the probability that \(S\), the number of heads in \(n\) tosses, lies within certain ranges.

\[\begin{align*} \mu &= n \times p = 1,000,000 \times 0.5 = 500,000 \\ \sigma &= \sqrt{n \times p \times (1-p)} = \sqrt{1,000,000 \times 0.5 \times 0.5} = \sqrt{250,000} = 500 \end{align*}\]

So, \(k = \frac{500}{500} = 1\).

According to Chebyshev’s inequality, the probability that \(S\) falls within \(499,500\) and \(500,500\) is at least \(1 - \frac{1}{1^2} = 0\).

The same as above, but now \(k = \frac{1000}{500} = 2\).

According to Chebyshev’s inequality, the probability that \(S\) falls within \(499,000\) and \(501,000\) is at least \(1 - \frac{1}{2^2} = 0.75\).

The same as above, but now \(k = \frac{1500}{500} = 3\).

According to Chebyshev’s inequality, the probability that \(S\) falls within \(498,500\) and \(501,500\) is at least \(1 - \frac{1}{3^2} = \frac{8}{9}\).

The Central Limit Theorem states that the distribution of the sum (or average) of a large number of independent, identically distributed random variables approaches a normal distribution, regardless of the original distribution. The mean of the distribution approaches \(\mu\), and the standard deviation approaches \(\frac{\sigma}{\sqrt{n}}\).

For all three intervals, we can use the CLT to approximate the probabilities using the normal distribution.

\[\begin{align*} \mu &= 500,000 \\ \sigma &= 500 \end{align*}\]

The width of each interval is 1000. So, we need to calculate the z-scores for the lower and upper bounds of each interval and then find the area under the normal curve between those z-scores.

Let’s use the z-score formula: \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value of the lower or upper bound of the interval.

For example, for the interval \(499,500 \leq S \leq 500,500\): \[\begin{align*} z_{\text{lower}} &= \frac{499,500 - 500,000}{500} = -1 \\ z_{\text{upper}} &= \frac{500,500 - 500,000}{500} = 1 \end{align*}\]

We can then look up the area under the standard normal curve between \(z_{\text{lower}}\) and \(z_{\text{upper}}\) to get the probability.

Similarly, we calculate for the other intervals. The probabilities can be calculated using a standard normal distribution table or using software.

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