Question

When you flip a coin, the probability of its landing on each side isin Equations 28-2 and 28-3. If you flip it times, the expected number of heads equals the expected number of tailsThe expected standard deviation for flips is. From Table 4-1, we expect thatof the results will lie withinand of the results will lie within.

(a) Find the expected standard deviation for the number of heads incoin flips.

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

(c) If you repeat thecoin flips many times, what is the expected range for the number of heads that includesof the results? (For example, your answer might be, "The rangetowill be observedof the time.")

Short answer

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

(b) The value of \(z\)is\(1.647\).

(c)The expected range for the number of heads is\(474 - 526\).

Step-by-step solution

Definition of Standard deviation.

• 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.

Determine the Standard deviation and the value of \[z\].

a)

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

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

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\(z\) :

\(\begin{array}{l}\frac{{z - 1.6}}{{1.7 - 1.6}} = \frac{{0.45 - 0.4452}}{{0.4554 - 0.4452}}\\z - 1.6 = \frac{{0.45 - 0.4452}}{{0.4554 - 0.4452}} \times (1.7 - 1.6)\\z = \left[ {\frac{{0.45 - 0.4452}}{{0.4554 - 0.4452}} \times (1.7 - 1.6)} \right] + 1.6\\z = 1.647\end{array}\)

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

Determine the expected range for the number of heads.

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:

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

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\(474 - 526\).