Chapter 5.3, Problem 69E

(a)

To determine

The values of n and p for X, which is binomially distributed.

Answer to Problem 69E

Solution: The required distribution of X is $\underset{\_}{p=0.62\text{ and }n=4}$.

Explanation of Solution

It is provided that 62% of the selected people admitted that they had received nasty messages on social sites. So, the probability of success for this case is $p=0.62$. The total number of people, which is 4, is randomly selected, so the total number of trails $n=4$. Assume X of them admitted that they had received nasty messages. Here X is binomially distributed with parameters $p=0.62\text{ and }n=4$. So, the required distribution of X is $p=0.62\text{ and }n=4$.

(b)

To determine

To find: The probability of every possible value of X.

Answer to Problem 69E

Solution: Probability distribution for each value of X is:

X	$P\left(X\right)$
0	0.02085
1	0.1360
2	0.333
3	0.3622
4	0.1477

Explanation of Solution

Calculation: In the provided problem the distribution can be calculated as:

$P\left(X\right)=\left(\begin{array}{l}n\\ X\end{array}\right)P{\left(1-P\right)}^{X-1}$

For $X=0$.

$\begin{array}{c}P\left(0\right)=\left(\begin{array}{l}4\\ 0\end{array}\right)\left(0.62\right){\left(1-0.62\right)}^4\\ =\left(\begin{array}{l}4\\ 0\end{array}\right)\left(0.62\right){\left(0.38\right)}^4\\ =0.02085\end{array}$

For $X=1$.

$\begin{array}{c}P\left(1\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)\left(0.62\right){\left(1-0.62\right)}^3\\ =\left(\begin{array}{l}4\\ 1\end{array}\right)\left(0.62\right){\left(0.38\right)}^3\\ =0.1360\end{array}$

For $X=2$.

$\begin{array}{c}P\left(2\right)=\left(\begin{array}{l}4\\ 2\end{array}\right)\left(0.62\right){\left(1-0.62\right)}^2\\ =\left(\begin{array}{l}4\\ 2\end{array}\right)\left(0.62\right){\left(0.38\right)}^2\\ =0.333\end{array}$

For $X=3$.

$\begin{array}{c}P\left(3\right)=\left(\begin{array}{l}4\\ 3\end{array}\right)\left(0.62\right){\left(1-0.62\right)}^1\\ =\left(\begin{array}{l}4\\ 3\end{array}\right)\left(0.62\right){\left(0.38\right)}^1\\ =0.3622\end{array}$

For $X=4$.

$\begin{array}{c}P\left(4\right)=\left(\begin{array}{l}4\\ 4\end{array}\right)\left(0.62\right){\left(1-0.62\right)}^0\\ =\left(\begin{array}{l}4\\ 4\end{array}\right)\left(0.62\right){\left(0.38\right)}^0\\ =0.1477\end{array}$

Hence, the table representing every possible value of X and the probability parallel to X is shown below:

X	$P\left(X\right)$
0	0.02085
1	0.1360
2	0.333
3	0.3622
4	0.1477

To determine

To graph: A probability histogram for the obtained distribution.

Explanation of Solution

Graph: In the provided problem, the distribution of X is obtained in the part above. Now, the probability histogram can be obtained by using Minitab. The steps to be followed are enlisted below:

Step 1: Input the values of X and the probability in Minitab worksheet.

Step 2: Go to Graph > Histogram.

Step 3: Choose ‘View simple’ and click OK.

Step 4: Insert the value of X and the probability p in the column of graph variables.

Step 5: Click on Ok.

The Histogram is obtained as:

The probability histogram is obtained for the required distribution of X.

(c)

To determine

To Find: The average number of responders who said yes.

Answer to Problem 69E

Solution: The required average is 2.48.

Explanation of Solution

Calculation: Binomial mean can be calculated using this formula,

${\mu }_x=np$

Substitute the values $p=0.62\text{ and }n=4$ in the formula above to obtain the mean. Mean can be calculated as:

$\begin{array}{c}{\mu }_x=np\\ =4\left(0.62\right)\\ =2.48\end{array}$

The required average is 2.48.

To determine

To graph: The average value in the obtained histogram.

Explanation of Solution

Graph: The obtained average value is 2.48. To mark this value in the obtained histogram, the steps are as follows:

Step 1: Go to the histogram which is obtained above.

Step 2: Double click on the X axis.

Step 3: Select ‘Position of ticks’ and enter the obtained mean value 2.48.

Step 4: Click on Ok.

The histogram obtained is:

The histogram with the mark of mean value 2.48 is obtained.