Chapter 5.CR, Problem 5CR

To determine

(a)

To find:

The standard deviation of this distribution and interpret it.

Answer to Problem 5CR

Solution:

The standard deviation tells how much the result varies from the mean value. So, the results may vary from the mean 2.5 by about 1.118 on average.

Explanation of Solution

Given:

Number of heads in five tosses of a coin.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

Where $x_i$ is the $i^{th}$ value of the random variable X.

The variance for a discrete probability distribution of random variable X is given by $\begin{array}{rll}{\sigma }^2& =& {\displaystyle \sum \left[x_i{}^2.P\left(X=x_i\right)\right]}-{\mu }^2\\ & =& {\displaystyle \sum \left[{\left(x_i-\mu \right)}^2.P(X=x_i)\right]}\end{array}$

Where $x_i$ is the $i^{th}$ value of the random variable X and $\mu$ is the mean of the probability distributions.

Calculation:

The possible combination of a coin which had tossed five times:

H	H	H	H	H
H	H	H	H	T
H	H	H	T	H
H	H	H	T	T
H	H	T	H	H
H	H	T	H	T
H	H	T	T	H
H	H	T	T	T
H	T	H	H	H
H	T	H	H	T
H	T	H	T	H
H	T	H	T	T
H	T	T	H	H
H	T	T	H	T
H	T	T	T	H
H	T	T	T	T
T	H	H	H	H
T	H	H	H	T
T	H	H	T	H
T	H	H	T	T
T	H	T	H	H
T	H	T	H	T
T	H	T	T	H
T	H	T	T	T
T	T	H	H	H
T	T	H	H	T
T	T	H	T	H
T	T	H	T	T
T	T	T	H	H
T	T	T	H	T
T	T	T	T	H
T	T	T	T	T

Total number of possible combinations are 32.

Let X denotes the number of heads.

The probability distribution table is shown below:

X	0	1	2	3	4	5
P(X = x)	$\frac{1}{32}$	$\frac{5}{32}$	$\frac{10}{32}$	$\frac{10}{32}$	$\frac{5}{32}$	$\frac{1}{32}$

The expected value can be calculated as follows:

$\begin{array}{rll}E(X)& =& {\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}\\ & =& \left[0\times \frac{1}{32}+1\times \frac{5}{32}+2\times \frac{10}{32}+3\times \frac{10}{32}+4\times \frac{5}{32}+5\times \frac{1}{32}\right]\\ & =& 2.5\end{array}$

The standard deviation is calculated as follows:

$\begin{array}{rll}\sigma & =& \sqrt{{\displaystyle \sum \left[x_i{}^2.P\left(X=x_i\right)\right]}-{\mu }^2}\\ & =& \sqrt{0^2\times \frac{1}{32}+1^2\times \frac{5}{32}+2^2\times \frac{10}{32}+3^2\times \frac{10}{32}+4^2\times \frac{5}{32}-{2.5}^2}\\ & =& 1.118\end{array}$

Therefore, the standard deviation tells how much the result varies from the mean value. So, the results may vary from the mean 2.5 by about 1.118 on average.

To determine

(b)

To find:

The standard deviation for the coin toss distribution and if it matches the one with calculated in part (a).

Answer to Problem 5CR

Solution::

The standard deviation for the coin toss distribution is 1.118 and it matches the one with calculated in part (a).

Explanation of Solution

Formula Used:

For a binomial distribution, the variance is given by ${\sigma }^2=np(1-p)$.

Calculation:

The probability of success in a coin toss is $\text{p=0}\text{.5}$, as only head and tail can appear in a coin toss with equal probabilities. The number of coin toss is $\text{n=5}$. The standard deviation of the binomial distribution is calculate as:

$\begin{array}{rll}\sigma & =& \sqrt{5\times 0.5(1-0.5)}\\ & =& \sqrt{5\times 0.5\times 0.5}\\ & =& 1.118\end{array}$

The standard deviation value matches the standard deviation value from part (a) because a binomial distribution is inherently a discrete probability distribution and the expected value is the mean of a binomial distribution. The standard deviation value indicated how the results will vary from the mean. Since the mean is the same as expected value, it makes sense that the standard deviation values of (a) and (b) matched.