Chapter 11.4, Problem 6CYU

a.

To determine

Determine the probabilities associated with the number of students.

Answer to Problem 6CYU

  $\boxed{0.37714952}$

Explanation of Solution

Given information:

A poll at steve’s high school was taken to see if students are in favour of spending class money to expand the junior-senior parking lot. Steve surveyed $6$ random students from the population.

Determine the probabilities associated with the number of students that Steve asked who are in favour of expanding the parking lot by calculating the probability distribution.

Calculation:

A poll was conducted in a school to know if the students are spending class money to expand the junior-senior parking lot. A boy named, S, in the school surveyed $6$ random students from the population.

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 $85\%$ of the students vote in favor, so,

  $\begin{array}{l}p=85\%\\ =0.85\end{array}$

While the remaining $15\%$ opposite it, so,

  $\begin{array}{l}q=15\%\\ =0.15\end{array}$

Substitute $p=0.85,q=0.15,k=0,n=6$ in the above formula to find the probability that none of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(0){=}^6{C}_0{(}^{0.85}{(}^{0.15}\\ =1(1){(}^{0.15}\\ =0.00001139\end{array}$

Hence, the probability that none of the $6$ random students surveyed by $S$ vote in favor is $\boxed{0.00001139}$ .

Substitute $p=0.85,q=0.15,k=1,n=6$ in the above formula to find the probability that one of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(1){=}^6{C}_1{(}^{0.85}{(}^{0.15}\\ =6(0.85){(}^{0.15}\\ =0.00038728\end{array}$

Hence, the probability that exactly one student of the 6 random students surveyed by S, vote in favor is

  $\boxed{0.00038728}$

Substitute $p=0.85,q=0.15,k=2,n=6$ in the above formula to find the probability that $2$ of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(2){=}^6{C}_2{(}^{0.85}{(}^{0.15}\\ =15{(}^{0.85}{(}^{0.15}\\ =0.00548648\end{array}$

Hence, the probability that exactly $2$ students of the $6$ random students surveyed by S, vote in favour is $\boxed{0.00548648}$

Substitute $p=0.85,q=0.15,k=3,n=6$ in the above formula to find the probability that $3$ of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(3){=}^6{C}_3{(}^{0.85}{(}^{0.15}\\ =20{(}^{0.85}{(}^{0.15}\\ =0.04145344\end{array}$

Hence, the probability that exactly $3$ students of the $6$ random students surveyed by S, vote in favor is $\boxed{0.04145344}$

Substitute $p=0.85,q=0.15,k=4,n=6$ in the above formula to find the probability that $4$ of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(4){=}^6{C}_4{(}^{0.85}{(}^{0.15}\\ =15{(}^{0.85}{(}^{0.15}\\ =0.17617711\end{array}$

Hence, the probability that exactly $4$ students of the $6$ random students surveyed by S, vote in favor is $\boxed{0.17617711}$

Substitute $p=0.85,q=0.15,k=5,n=6$ in the above formula to find the probability that $5$ of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(5){=}^6{C}_5{(}^{0.85}{(}^{0.15}\\ =6{(}^{0.85}{(}^{0.15}\\ =0.39933478\end{array}$

Hence, the probability that exactly $5$ students of the $6$ random students surveyed by S, vote in favor is $\boxed{0.39933478}$

Substitute $p=0.85,q=0.15,k=6,n=6$ in the above formula to find the probability that $6$ of the $6$ random students surveyed by $S$ vote in favor,

  $\begin{array}{l}p(6){=}^6{C}_6{(}^{0.85}{(}^{0.15}\\ =1{(}^{0.85}{(}^{0.15}\\ =0.37714952\end{array}$

Hence, the probability that all $6$ random students surveyed by S, vote in favor is $\boxed{0.37714952}$

b.

To determine

What is the probability that no more than $2$ people are in favour of expanding the parking lot?

Answer to Problem 6CYU

  $\boxed{0.00588067}$

Explanation of Solution

Given information:

A poll at steve’s high school was taken to see if students are in favour of spending class money to expand the junior-senior parking lot. Steve surveyed $6$ random students from the population.

What is the probability that no more than $2$ people are in favour of expanding the parking lot?

Calculation:

A poll was conducted in a school to know if the students are spending class money to expand the junior-senior parking lot. A boy named, S, in the school surveyed $6$ random students from the population.

The number of people out of $6$ random people who vote in favor could be $0,1,or2$ .

From part a, it is clear that,

  $\begin{array}{l}P(0)=0.00001139\\ P(0)=0.00038728\end{array}$

  $P(2)=0.00548648$

So, the probability that no mora than $2$ out of $6$ random people surveyed vote in favor of expanding the parking lot is,

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

  $\begin{array}{l}=0.00001139+0.00038728+0.00548648\\ =0.00588067\end{array}$

Hence, the probability that no more than $2$ out of 6 random people surveyed vote in favor of expanding the parking lot is $\boxed{0.00588067}$

c.

To determine

How many students should Steve expect to find who are in favour of expanding the parking lot?

Answer to Problem 6CYU

  $\boxed{5}$

Explanation of Solution

Given information:

A poll at steve’s high school was taken to see if students are in favour of spending class money to expand the junior-senior parking lot. Steve surveyed $6$ random students from the population.

How many students should Steve expect to find who are in favour of expanding the parking lot?

Calculation:

A poll was conducted in a school to know if the students are spending class money to expand the junior-senior parking lot. A boy named, S, in the school surveyed $6$ random students from the population.

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 $n=6,p=0.85$ . Substitute these values in the above formula to find the number of students that the boy named S shuold expect to find who vote in favor of expanding the parking lot,

  $\begin{array}{l}\mu =np\\ =6(0.85)\\ =5.1\end{array}$

  $\approx 5$

Hence, the number of students that the boy named S should expect to find who vote in favor of expanding the parking lot is $\boxed{5}$ .