Chapter 10, Problem 10CT

a.

To determine

Tograph:

The histogram of binomial distribution.

Explanation of Solution

Given information:

Family dinner at least four time per week $=58\%$

Randomly chooses family $=5$

Graph:

  

Interpretation:

In the above histogram, the probability of binomial distribution of the survey is described.

For a binomial experiment the probability of exactly, $k$ successes in $n$ trials is as follows:

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

Where,

  $p$ is probability of success.

From information given,

  $n=5,p=0.58,k=0,1,2,\mathrm{...5.}$

Now, substitution the values in the stated formula and obtain the required probability. The required probabilities are as follows:

  $\begin{array}{l}P\left(k=0\right){=}_5{C}_0{\left(0.58\right)}^0{\left(0.42\right)}^5\\ =0.0131\\ P\left(k=1\right){=}_5{C}_1{\left(0.58\right)}^1{\left(0.42\right)}^4\\ =0.0902\\ P\left(k=2\right){=}_5{C}_2{\left(0.58\right)}^2{\left(0.42\right)}^3\\ =0.2492\\ P\left(k=3\right){=}_5{C}_3{\left(0.58\right)}^3{\left(0.42\right)}^2\\ =0.3442\end{array}$

Equivalently, $P\left(k=4\right)$ is $0.2376$ , $P\left(k=5\right)$ is $0.0656$ .

b.

To determine

To calculate:

The most likely outcome of the survey.

Answer to Problem 10CT

The most likely outcome of the survey is $3$ .

Explanation of Solution

Given information:

Family dinner at least four time per week $=58\%$

Randomly chooses family $=5$

Calculation:

The most likely outcome of the survey is the value of survey is the value of $x$ for which $P\left(x\right)$ is highest.

The probability distribution of the survey results show that the probability is greatest for $3$ , nearly $0.34$ .

Therefore, the most likely outcome of the survey is $3$ .

c.

To determine

To calculate:

The probability of at least $3$ families having dinner four times per week.

Answer to Problem 10CT

The probability of at least $3$ families having dinner four times per week is $0.6475$ .

Explanation of Solution

Given information:

Family dinner at least four time per week $=58\%$

Randomly chooses family $=5$

Calculation:

The required probability is as follows:

  $\begin{array}{l}P\left(k\ge 3\right)=P\left(k=3\right)+P\left(k=4\right)+P\left(k=5\right)\\ =0.3442+0.2376+0.0657\\ =0.6475\end{array}$

Hence, the probability of at least $3$ families having dinner four times per week is $0.6475$ .