Chapter 8.3, Problem 8.60E

(a)

To determine

To calculate the expected number of dog owners who greet their dog first when they arrive home.

Answer to Problem 8.60E

  $7.92$ .

Explanation of Solution

It follows binomial distribution.

  $\begin{array}{l}n=12\\ p=0.66\end{array}$

The expected value is:

  $\begin{array}{l}EV(X)=np\\ =12(0.66)\\ =7.92\end{array}$

(b)

To determine

To make a table that shows the probability distribution of X = the number of dog owners who greet their dog first when they arrive home.

Answer to Problem 8.60E

  

Explanation of Solution

  $\begin{array}{l}n=12\\ p=0.66\end{array}$

Evaluate at $k=0,1,2,3,4,5,6,7,8,9,10,11,12$ :

  $P(X=0)=C{}_0\times {(0.66)}^0\times {(1-0.66)}^{12-0}=2.38642\times {10}^{-6}$

  $P(X=1)=C{}_1\times {(0.66)}^1\times {(1-0.66)}^{12-1}=0.0000556$

  $P(X=2)=C{}_2\times {(0.66)}^2\times {(1-0.66)}^{12-2}=0.0005935$

  $P(X=3)=C{}_3\times {(0.66)}^3\times {(1-0.66)}^{12-3}=0.00384$

  $P(X=4)=C{}_4\times {(0.66)}^4\times {(1-0.66)}^{12-4}=0.01677$

  $P(X=5)=C{}_5\times {(0.66)}^5\times {(1-0.66)}^{12-5}=0.0521$

  $P(X=6)=C{}_6\times {(0.66)}^6\times {(1-0.66)}^{12-6}=0.1180$

  $P(X=7)=C{}_7\times {(0.66)}^7\times {(1-0.66)}^{12-7}=0.1963$

  $P(X=8)=C{}_8\times {(0.66)}^8\times {(1-0.66)}^{12-8}=0.2382$

  $P(X=9)=C{}_9\times {(0.66)}^9\times {(1-0.66)}^{12-9}=0.2055$

  $P(X=10)=C{}_{10}\times {(0.66)}^{10}\times {(1-0.66)}^{12-10}=0.1197$

  $P(X=11)=C{}_{11}\times {(0.66)}^{11}\times {(1-0.66)}^{12-11}=0.0422$

  $P(X=12)=C{}_{12}\times {(0.66)}^{12}\times {(1-0.66)}^{12-12}=0.006832$

The table is as:

  

(c)

To determine

To construct a histogram that shows the probability distribution of X and describe.

Explanation of Solution

The histogram is:

  

We note that the probability histogram is left skewed because the highest bars are to the right in the histogram.