Chapter 10.6, Problem 17E

(a)

To determine

To calculate:

The probability value of $x$

Answer to Problem 17E

The probability value of $x$ is

  $\begin{array}{l}P\left(k=0\right)=0.099,P\left(k=1\right)=0.271,P\left(k=2\right)=0.319,P\left(k=3\right)=0.208,P\left(k=4\right)=0.081,P\left(k=5\right)=0.081,\\ P\left(k=6\right)=0.081,P\left(k=7\right)=0.081\end{array}$

Explanation of Solution

Given information:

  

The probability value of: $x=0,1,2,\cdot \cdot \cdot ,7$

  $$

Calculation:

The binomial experiment, the probability of k successes in number of trial is:

  $P\left(ksuccesses\right)=C{}_k{p}^k{\left(1-p\right)}^{n-k}\mathrm{...}\left(1\right)$

Put the values $n=7,p=0.28125,k=0,1,2,\cdot \cdot \cdot ,7$ in equation (1):

  $\begin{array}{l}P\left(k=0\right)=C{}_0{\left(0.28125\right)}^0{\left(0.71875\right)}^7\\ P\left(k=0\right)=0.099\\ P\left(k=1\right)=C{}_1{\left(0.28125\right)}^1{\left(0.71875\right)}^6\\ P\left(k=1\right)=0.271\\ P\left(k=2\right)=C{}_2{\left(0.28125\right)}^2{\left(0.71875\right)}^5\\ P\left(k=2\right)=0.319\\ P\left(k=3\right)=C{}_3{\left(0.28125\right)}^3{\left(0.71875\right)}^4\\ P\left(k=3\right)=0.208\\ P\left(k=4\right)=C{}_4{\left(0.28125\right)}^4{\left(0.71875\right)}^3\\ P\left(k=4\right)=0.081\\ P\left(k=5\right)=C{}_5{\left(0.28125\right)}^5{\left(0.71875\right)}^2\\ P\left(k=5\right)=0.081\\ P\left(k=6\right)=C{}_6{\left(0.28125\right)}^6{\left(0.71875\right)}^1\\ P\left(k=6\right)=0.081\\ P\left(k=7\right)=C{}_7{\left(0.28125\right)}^7{\left(0.71875\right)}^0\end{array}$

The probability value of $x$ is

  $\begin{array}{l}P\left(k=0\right)=0.099,P\left(k=1\right)=0.271,P\left(k=2\right)=0.319,P\left(k=3\right)=0.208,P\left(k=4\right)=0.081,P\left(k=5\right)=0.081,\\ P\left(k=6\right)=0.081,P\left(k=7\right)=0.081\end{array}$

(b)

To determine

To calculate:

To make the table showing the probability distribution for $x$

Answer to Problem 17E

The table showing the probability distribution:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P\left(x\right)$	$0.099$	$0.271$	$0.319$	$0.208$	$0.081$	$0.019$	$0.0025$	0.00014

Explanation of Solution

Given information:

  

Calculation:

The probability value of $x$ is follows using sub-part (a)

  $\begin{array}{l}P\left(k=0\right)=0.099,P\left(k=1\right)=0.271,P\left(k=2\right)=0.319,P\left(k=3\right)=0.208,P\left(k=4\right)=0.081,P\left(k=5\right)=0.081,\\ P\left(k=6\right)=0.081,P\left(k=7\right)=0.081\end{array}$

The table showing the probability distribution:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P\left(x\right)$	$0.099$	$0.271$	$0.319$	$0.208$	$0.081$	$0.019$	$0.0025$	0.00014

(c)

To determine

To graph:

The histogram showing the probability distribution for $x$

Answer to Problem 17E

The histogram of the probability distribution:

  

Graph: Histogram

Explanation of Solution

Given information:

  

Interpretation:

  

Graph: Histogram

The histogram of the probability distribution is shown as above.