Chapter 11.4, Problem 22PPS

a.

To determine

Determine the probabilities associated with the number of students.

Answer to Problem 22PPS

  $\boxed{0.000064or0.0064\%}$ , $\boxed{0.001536or0.1536\%}$ , $\boxed{0.015360or0.5360\%}$ , $\boxed{0.081920or8.1920\%}$ , $\boxed{0.245760or24.5760\%}$ , $\boxed{0.393216or39.3216\%}$ and $\boxed{0.262144or26.2144\%}$

Explanation of Solution

Given information:

A survey was taken to see the percent of students that participate in sports for their school. Six random students are chosen.

Determine the probabilities associated with the number of students playing in at least one sport by calculating the probability distribution.

Calculation:

A survey finds that $20\%$ of do not participate in sports for their, $55\%$ participate in exactly $1$ sport, $20\%$ participate in exactly $2$ sports, and $5\%$ participate in $3$ or more that $3$ sports for their school. Suppose that $6$ school students are selected at random and they are asked how many sports they participate for their school.

The percentage of students participating in at least one sport for their school is, $100-20=80$

So, it is clear that $80\%$ of students in a school participate in at least one sport for their school.

The probability of $k$ successes in $n$ trials of a binomial experiment is,

  $P(k){=}^n{C}_k{(p)}^k{(q)}^{n-k}$

Here $p$ represents the probability of success and $q$ represents the probability of failure.

Since $80\%$ of the students in a school participate in at least one sport for their school, so,

  $\begin{array}{l}p=80\%\\ =0.8\end{array}$

The probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.8\\ =0.2\end{array}$

Substitute $p=0.8,q=0.2,k=0,n=6$ in the above formula to find the probability that none of the $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(0){=}^6{C}_0{(}^{0.8}{(}^{0.2}\\ =1(1){(}^{0.2}\\ =0.000064\\ =0.0064\%\end{array}$

Hence, the probability that none of the $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.000064or0.0064\%}$ .

Substitute $p=0.8,q=0.2,k=1,n=6$ in the above formula to find the probability that exactly one of the $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(1){=}^6{C}_1{(}^{0.8}{(}^{0.2}\\ =6(0.8){(}^{0.2}\\ =0.001536\\ =0.1536\%\end{array}$

Hence, the probability that exactly $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.001536or0.1536\%}$ .

Substitute $p=0.8,q=0.2,k=2,n=6$ in the above formula to find the probability that exactly 2of the $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(2){=}^6{C}_2{(}^{0.8}{(}^{0.2}\\ =15{(}^{0.8}{(}^{0.2}\\ =0.015360\\ =0.5360\%\end{array}$

Hence, the probability that exactly $2$ of the $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.015360or0.5360\%}$ .

Substitute $p=0.8,q=0.2,k=3,n=6$ in the above formula to find the probability that exactly $3$ of the $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(3){=}^6{C}_3{(}^{0.8}{(}^{0.2}\\ =20{(}^{0.8}{(}^{0.2}\\ =0.081920\\ =8.1920\%\end{array}$

Hence, the probability that exactly $3$ of the $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.081920or8.1920\%}$ .

Substitute $p=0.8,q=0.2,k=4,n=6$ in the above formula to find the probability that exactly $4$ of the $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(4){=}^6{C}_4{(}^{0.8}{(}^{0.2}\\ =15{(}^{0.8}{(}^{0.2}\\ =0.245760\\ =24.5760\%\end{array}$

Hence, the probability that exactly $4$ of the $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.245760or24.5760\%}$ .

Substitute $p=0.8,q=0.2,k=5,n=6$ in the above formula to find the probability that exactly $5$ of the $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(5){=}^6{C}_5{(}^{0.8}{(}^{0.2}\\ =6{(}^{0.8}{(}^{0.2}\\ =0.393216\\ =39.3216\%\end{array}$

Hence, the probability that exactly $5$ of the $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.393216or39.3216\%}$ .

Substitute $p=0.8,q=0.2,k=6,n=6$ in the above formula to find the probability that all $6$ random students surveyed participate in at least one sport for their school,

  $\begin{array}{l}p(6){=}^6{C}_6{(}^{0.8}{(}^{0.2}\\ =6{(}^{0.8}\\ =0.262144\\ =26.2144\%\end{array}$

Hence, the probability that all $6$ random students surveyed participate in at least one sport for thair school is $\boxed{0.262144or26.2144\%}$ .

b.

To determine

What is the probability that no more than $2$ of the students participated in a sport?

Answer to Problem 22PPS

  $\boxed{0.083520or8.3520\%}$

Explanation of Solution

Given information:

A survey was taken to see the percent of students that participate in sports for their school. Six random students are chosen.

What is the probability that no more than $2$ of the students participated in a sport?

Calculation:

Consider the probability that no more $2$ of the $6$ random sudents participated in a sport.

The number of students out of $6$ random students who partcipated in a spot could be $0,1or2$

From part a, it is clear that,

  $\begin{array}{l}P(0)=0.000064\\ P(1)=0.001536\\ P(2)=0.081920\end{array}$

Hence, the probability that at no more $2$ of the $6$ random sudents participated in a sport is,

  $\text{P}\left(\text{No more than 2}\right)=\text{P}\left(0\right)+\text{P}\left(\text{1}\right)+\text{P}\left(\text{2}\right)$

  $\begin{array}{l}=0.000064+0.001536+0.081920\\ =0.083520\\ =8.3520\%\end{array}$

Hence, the probability that no more $2$ of the $6$ random sudents participated in a sport is

  $\boxed{0.083520or8.3520\%}$ .

c.

To determine

How many students should you expect to have participated in at least one sport?

Answer to Problem 22PPS

  $\boxed{5}$

Explanation of Solution

Given information:

A survey was taken to see the percent of students that participate in sports for their school. Six random students are chosen.

How many students should you expect to have participated in at least one sport?

Calculation:

Generally, the mean $\mu$ , of any binomial distribution of success.

  $\mu =np$

Here $n$ is the number of trials and $p$ is the probability of success.

For the above situation, it is clear that $n=10,p=0.8$ . Substitute these values in the above formula to find the number of students expected to to participate in at least one sport ,

  $\begin{array}{l}\mu =np\\ =6(0.8)\\ =4.8\end{array}$

  $\approx 5$

Hence, the number of students who are expected to participate in at least one sport $\boxed{5}$ .