Chapter 5.2, Problem 59E

(a)

To determine

To find: The value of n and p in the binomial distribution of X.

Answer to Problem 59E

Solution: $n=4,p=0.7$.

Explanation of Solution

Calculation: The formula of the binomial distributions is:

$P\left(X\right)=C{}_X{p}^X{\left(1-p\right)}^{n-X}$

Here, n is number of trials indefinitely and p is the probability that success.

So, based on the provided information, the success probability is the probability that the person responds in favor of the research question. And the number of independent trails are the randomly selected persons.

So, $n=4$ and $p=0.7$.

Interpretation: The number of success trials $\left(n\right)$ is 4 and the success probability $\left(p\right)$ is $0.7$.

(b)

To determine

To find: The probability of each possible outcomes of X, and the histogram for the distribution

Answer to Problem 59E

Solution: The probability of each possible value of X is:

X	$P\left(X\right)$
0	$0.0081$
1	$0.0756$
2	$0.2646$
3	$0.4116$
4	$0.2401$

Explanation of Solution

Calculation: The probability for $X=0$ at $n=4$ and $p=0.7$ is calculated as:

$\begin{array}{c}P\left(X=0\right)=C{}_0\times {0.7}^0{\left(1-0.7\right)}^{4-0}\\ =1\times 1\times {\left(0.3\right)}^4\\ =0.0081\end{array}$

Similarly, the probability for $X=1$ is,

$\begin{array}{c}P\left(X=1\right)=C{}_1\times {0.7}^1{\left(1-0.7\right)}^{4-1}\\ =4\times 0.7\times {\left(0.3\right)}^3\\ =0.0756\end{array}$

Similarly, the probability for $X=2$ is,

$\begin{array}{c}P\left(X=2\right)=C{}_2\times {0.7}^2{\left(1-0.7\right)}^{4-2}\\ =6\times {0.7}^2\times {\left(0.3\right)}^2\\ =0.2646\end{array}$

Similarly, the probability for $X=3$ is,

$\begin{array}{c}P\left(X=3\right)=C{}_3\times {0.7}^3{\left(1-0.7\right)}^{4-3}\\ =4\times {0.7}^3\times {\left(0.3\right)}^1\\ =0.4116\end{array}$

Similarly, the probability for $X=4$ is,

$\begin{array}{c}P\left(X=4\right)=C{}_4\times {0.7}^4{\left(1-0.7\right)}^{4-4}\\ =1\times {0.7}^4\times {\left(0.3\right)}^0\\ =0.2401\end{array}$

Therefore, the probability of each possible value of X is,

X	$P\left(X\right)$
0	$0.0081$
1	$0.0756$
2	$0.2646$
3	$0.4116$
4	$0.2401$

Graph: The graph of the probability histogram makes by use of the follows steps:

Step 1: Put the data in the excel sheet.

X	$P\left(X\right)$
0	$0.0081$
1	$0.0756$
2	$0.2646$
3	$0.4116$
4	$0.2401$

Step 2: Select the data set and go to insert and select the option of cluster column under the Recommended Charts. The screenshot is shown below:

Step 3: Click on OK. The diagram is obtained as:

Step 4: Click on the chart area and select the option of “Primary Horizontal” and “Primary Vertical” axis under the “Add Chart Element” to add the axis title. The screenshot is shown below:

Step 5: Click on the bars of the diagrams and reduce the gap width to zero under the “Format Data Series” tab. The screenshot is shown below:

The obtained histogram is:

Interpretation: The probability $0.4116$ is maximum for the success trials 3 and the histogram indicates that the distribution is approximately symmetric.

(c)

To determine

To find: The mean number of positive responders and mark the location of this value on your histogram.

Answer to Problem 59E

Solution: The mean $\left(\mu \right)$ is $2.8$.

Explanation of Solution

Calculation: The number of success trials $\left(n\right)$ is 4 and the probability $\left(p\right)$ is $0.7$ for the 4 success trials.

So, the mean $\left(\mu \right)$ of positive response is computed as:

$\begin{array}{c}\mu =np\\ =4\times 0.7\\ =2.8\end{array}$

Graph: The graph of the probability histogram makes by use of the follows steps:

Step 1: Put the data in the excel sheet.

X	$P\left(X\right)$
0	$0.0081$
1	$0.0756$
2	$0.2646$
3	$0.4116$
4	$0.2401$

Step 2: Select the data set and go to insert and select the option of cluster column under the Recommended Charts. The screenshot is shown below:

Step 3: Click on OK. The diagram is obtained as:

Step 4: Click on the chart area and select the option of “Primary Horizontal” and “Primary Vertical” axis under the “Add Chart Element” to add the axis title. The screenshot is shown below:

Step 5: Click on the bars of the diagrams and reduce the gap width to zero under the “Format Data Series” tab. The screenshot is shown below:

Step 6: The mean of 2.8 is marked on the histogram and the obtained histogram is:

Interpretation: The mean $\left(\mu \right)$ is $2.8$ which lies near the center of the histogram.