Chapter 11.4, Problem 23PPS

a.

To determine

Determine the probabilities.

Answer to Problem 23PPS

  $\boxed{0.0012or0.12\%}$ , $\boxed{0.0124or1.24\%}$ , $\boxed{0.0575or5.75\%}$ , $\boxed{0.1525or15.25\%}$ , $\boxed{0.2526or25.26\%}$ , $\boxed{0.2679or26.79\%}$ , $\boxed{0.1776or17.76\%}$ , $\boxed{0.0672or6.72\%}$ and $\boxed{0.0111or1.11\%}$.

Explanation of Solution

Given information:

An online poll showed that $57\%$ of adults still own vinyl records. Moe surveyed $8$ random adults from the population.

Determine the probabilities associated with the number of adults that still own vinyl records by calculating the probabilities distribution.

Calculation:

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 $57\%$ of the adults in a population own vinyl records, so,

  $\begin{array}{l}p=57\%\\ =0.57\end{array}$

The probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.57\\ =0.43\end{array}$

Substitute $p=0.57,q=0.43,k=0,n=8$ in the above formula to find the probability that none of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(0){=}^8{C}_0{(}^{0.57}{(}^{0.43}\\ =1(1){(}^{0.43}\\ =0.0012\\ =0.12\%\end{array}$

Hence, the probability that none of the $8$ random students surveyed own vinyl record is $\boxed{0.0012or0.12\%}$ .

Substitute $p=0.57,q=0.43,k=1,n=8$ in the above formula to find the probability that exactly one of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(1){=}^8{C}_1{(}^{0.57}{(}^{0.43}\\ =8(0.57){(}^{0.43}\\ =0.0124\\ =1.24\%\end{array}$

Hence, the probability that exactly one of the $8$ random students surveyed own vinyl record is $\boxed{0.0124or1.24\%}$ .

Substitute $p=0.57,q=0.43,k=2,n=8$ in the above formula to find the probability that exactly $2$ of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(2){=}^8{C}_2{(}^{0.57}{(}^{0.43}\\ =28{(}^{0.57}{(}^{0.43}\\ =0.0575\\ =5.75\%\end{array}$

Hence, the probability that exactly $2$ of the $8$ random students surveyed own vinyl record is $\boxed{0.0575or5.75\%}$ .

Substitute $p=0.57,q=0.43,k=3,n=8$ in the above formula to find the probability that exactly $3$ of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(3){=}^8{C}_3{(}^{0.57}{(}^{0.43}\\ =56{(}^{0.57}{(}^{0.43}\\ =0.1525\\ =15.25\%\end{array}$

Hence, the probability that exactly $3$ of the $8$ random students surveyed own vinyl record is $\boxed{0.1525or15.25\%}$ .

Substitute $p=0.57,q=0.43,k=4,n=8$ in the above formula to find the probability that exactly $4$ of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(4){=}^8{C}_4{(}^{0.57}{(}^{0.43}\\ =70{(}^{0.57}{(}^{0.43}\\ =0.2526\\ =25.26\%\end{array}$

Hence, the probability that exactly $4$ of the $8$ random students surveyed own vinyl record is $\boxed{0.2526or25.26\%}$ .

Substitute $p=0.57,q=0.43,k=5,n=8$ in the above formula to find the probability that exactly $5$ of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(5){=}^8{C}_5{(}^{0.57}{(}^{0.43}\\ =56{(}^{0.57}{(}^{0.43}\\ =0.2679\\ =26.79\%\end{array}$

Hence, the probability that exactly $5$ of the $8$ random students surveyed own vinyl record is $\boxed{0.2679or26.79\%}$ .

Substitute $p=0.57,q=0.43,k=6,n=8$ in the above formula to find the probability that exactly $6$ of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(6){=}^8{C}_6{(}^{0.57}{(}^{0.43}\\ =28{(}^{0.57}{(}^{0.43}\\ =0.1776\\ =17.76\%\end{array}$

Hence, the probability that exactly $6$ of the $8$ random students surveyed own vinyl record is $\boxed{0.1776or17.76\%}$ .

Substitute $p=0.57,q=0.43,k=7,n=8$ in the above formula to find the probability that exactly $7$ of the $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(7){=}^8{C}_7{(}^{0.57}{(}^{0.43}\\ =8{(}^{0.57}{(}^{0.43}\\ =0.0672\\ =6.72\%\end{array}$

Hence, the probability that exactly $7$ of the $8$ random students surveyed own vinyl record is $\boxed{0.0672or6.72\%}$ .

Substitute $p=0.57,q=0.43,k=8,n=8$ in the above formula to find the probability that all $8$ random adults surveyed own vinyl records,

  $\begin{array}{l}p(8){=}^8{C}_8{(}^{0.57}{(}^{0.43}\\ =1{(}^{0.57}(1)\\ =0.0111\\ =1.11\%\end{array}$

Hence, the probability that all $8$ random students surveyed own vinyl record is $\boxed{0.0111or1.11\%}$ .

b.

To determine

What is the probability that no less than $6$ of the people surveyed still own vinyl records?

Answer to Problem 23PPS

  $\boxed{0.2559or25.59\%}$

Explanation of Solution

Given information:

An online poll showed that $57\%$ of adults still own vinyl records. Moe surveyed $8$ random adults from the population.

What is the probability that no less than $6$ of the people surveyed still own vinyl records?

Calculation:

Consider the probability that no less than $6$ of the 8 random adults surveyed own vinyl records.

The number of adults out of 8 random students who own vinyl records could be $6,7or8$ .

From part a, it is clear that,

  $\begin{array}{l}P(6)=0.1776\\ P(1)=0.0672\\ P(8)=0.0111\end{array}$

Hence, the probability that at no less than $6$ of the 8 random adults surveyed own vinyl records is,

  $\text{P}\left(\text{No less than 6}\right)=\text{P}\left(\text{6}\right)+\text{P}\left(\text{1}\right)+\text{P}\left(\text{8}\right)$

  $\begin{array}{l}=0.1776+0.0672+0.0111\\ =0.2559\\ =25.59\%\end{array}$

Hence, the probability that no less than $6$ of the 8 random adults surveyed own vinyl records is,

  $\boxed{0.2559or25.59\%}$ .

c.

To determine

How many people should Moe expect to still own vinyl records?

Answer to Problem 23PPS

  $\boxed{5}$

Explanation of Solution

Given information:

An online poll showed that $57\%$ of adults still own vinyl records. Moe surveyed $8$ random adults from the population.

How many people should Moe expect to still own vinyl records?

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 $n=8,p=0.57$ . Substitute these values in the above formula to find the expected number of adults to own vinyl records.

  $\begin{array}{l}\mu =np\\ =8(0.57)\\ =4.56\end{array}$

  $\approx 5$

Hence, the expected number of adults to own vinyl records is $\boxed{5}$ .