A balanced coin is flipped 400 times. Determine the number x such that the probability that the number of heads is between \(200 − x\) and \(200 + x\) is approximately .80.
Let’s first define \(n=400\) as the number of trials, \(p=0.5\) as the probability of getting a head, and \(q=1-p=0.5\) as the probability of getting a tail.
We know that an approximation of the binomial probability is:
\[P(i \leq S_n \leq j) \approx NA \bigg(\frac{i- \frac{1}{2} - np}{\sqrt{npq}}, \frac{j + \frac{1}{2} - np}{\sqrt{npq}} \bigg)\]
We will use this to solve for \(x\) by substituting the lower and upper bounds in for \(i\) and \(j\) in our equation:
\[P(200-x \leq S_n \leq 200+x) \approx NA \bigg(\frac{ (200-x) - \frac{1}{2} - np}{\sqrt{npq}}, \frac{ (200+x) + \frac{1}{2} - np}{\sqrt{npq}} \bigg)\]
\[0.8\approx NA \bigg(\frac{ (200-x) - \frac{1}{2} - (400 \cdot 0.5)}{\sqrt{(400 \cdot 0.5 \cdot 0.5)}}, \frac{ (200+x) + \frac{1}{2} - (400 \cdot 0.5)}{\sqrt{(400 \cdot 0.5 \cdot 0.5)}} \bigg)\]
\[0.8 \approx NA \bigg(\frac{ 200-x - \frac{1}{2} - 200}{\sqrt{100}}, \frac{ 200+x + \frac{1}{2} - 200}{\sqrt{100}} \bigg)\]
\[0.8 \approx NA \bigg(\frac{ -x - \frac{1}{2}}{10}, \frac{ x + \frac{1}{2}}{10} \bigg)\]
\[0.8 \approx 2NA \bigg(\frac{ x + \frac{1}{2}}{10} \bigg) \]
\[0.4 \approx NA \bigg(\frac{ x + \frac{1}{2}}{10} \bigg) \]
We can look this value up in the table of values of \(NA(0, z)\), the normal area from \(0\) to \(z\) and solve:
\[1.3 = \frac{ x + \frac{1}{2}}{10}\]
\[12.5 = x \]