When you flip a coin, the probability of its landing on each side is \(p = q = \frac{1}{2}\)in Equations 28-2 and 28-3. If you flip it \(n\)times, the expected number of heads equals the expected number of tails \( = np = nq = \frac{1}{2}n.\)The expected standard deviation for \(n\)flips is\({\sigma _n} = \sqrt {npq} \). From Table 4-1, we expect that \(68.3\% \)of the results will lie within \( \pm 1{\sigma _n}\) and \(95.5\% \)of the results will lie within\( \pm 2{\sigma _n}\).

(a) Find the expected standard deviation for the number of heads in \({\bf{1000}}\) coin flips.

(b) By interpolation in Table 4-1, find the value of \(z\)that includes \(90\% \)of the area of the Gaussian curve. We expect that \(90\% \)of the results will lie within this number of standard deviations from the mean.

(c) If you repeat the\({\bf{1000}}\)coin flips many times, what is the expected range for the number of heads that includes\(90\% \) of the results? (For example, your answer might be, "The range \({\bf{490}}\) to \({\bf{510}}\) will be observed \(90\% \)of the time.")

• The standard deviation is a measure of how far something deviates from the mean (for example, spread, dispersion, or spread). A "typical" variation from the mean is represented by the standard deviation.
• Because it returns to the data set's original units of measurement, it's a common measure of variability.
• The standard deviation, defined as the square root of the variance, is a statistic that represents the dispersion of a dataset relative to its mean.

a)

Find the expected standard deviation for the number of heads in \(1000\) coin flips:

\(\begin{aligned}{}{\delta _n} &= \sqrt {npq} \\{\delta _n} &= \sqrt {{{10}^3} \times 0.5 \times 0.5} \\{\delta _n} &= 15.8\end{aligned}\)

Therefore, the standard deviation for the number of heads in \(1000\) coin flips is \(15.8\)

b)

From Table \(4 - 1\) we will find the value of \(z\) which includes \(90\% \) of the area of the Gaussian curve:

From Table \(4 - 1\) we can see that \(z\) area is $0.45$ (considering that area \({z^ - }\)to \({z^ + } = 0.90\) )

The value would lie between \(z = 1.6\) and \(z = 1.7\) with areas \(0.4452\)and \(0.4554\)

Next we will make the linear interpolation and calculate the value of \(z\) :

\(\begin{aligned}{}\frac{{z - 1.6}}{{1.7 - 1.6}} &= \frac{{0.45 - 0.4452}}{{0.4554 - 0.4452}}\\z - 1.6 &= \frac{{0.45 - 0.4452}}{{0.4554 - 0.4452}} \times (1.7 - 1.6)\\z &= \left( {\frac{{0.45 - 0.4452}}{{0.4554 - 0.4452}} \times (1.7 - 1.6)} \right) + 1.6\\z &= 1.647\end{aligned}\)

Hence, the value of \(z\)is \(1.647\).

c)

State the expected range for the number of heads that includes \(90\% \) of the results if we repeat the \(1000\)coin flips many times:

First consider the following

\(z = \frac{{(x - \bar x)}}{s},\quad x = \bar x \pm zs\)

Next calculate the expected range for the number of heads:

\(\begin{aligned}{}x &= \bar x \pm zs\\x &= 500 \pm (1.647 \times 15.8)\\x &= 500 \pm 26\end{aligned}\)

The range would be \(474 - 526\) considering that \(500 - 26 = 474\) and \(500 + 26 = 526\)

Also note that \(\bar x = 500\) because there are two sides on the coin (one is head and other is tail)

Therefore, the expected range for the number of heads is \(474 - 526\).