Chapter 11.4, Problem 21PPS

a.

To determine

Determine the probabilities associated with the number of students.

Answer to Problem 21PPS

  $\boxed{0.0006or0.06\%}$ , $\boxed{0.0007or0.07\%}$ , $\boxed{0.0343or3.43\%}$ , $\boxed{0.0991or9.91\%}$ , $\boxed{0.1878or18.78\%}$ , $\boxed{0.2441or24.41\%}$ , $\boxed{0.2204or22.04\%}$ , $\boxed{0.1364or13.64\%}$ , $\boxed{0.0554or5.54\%}$ , $\boxed{0.0133or1.33\%}$ , $\boxed{0.0014or1.14\%}$ .

Explanation of Solution

Given information:

According to a recent survey, $52\%$ of high school students own a laptop. Ten random students are chosen.

Determine the probability associated with the number of students that own a laptop by calculating the probability distribution.

Calculation:

A survey finds that $52\%$ of high school students own a laptop.

Suppose that $10$ high school students are selected at random and they are asked if they own a laptop.

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 $52\%$ of high school students own a laptop, so,

  $\begin{array}{l}p=52\%\\ =0.52\end{array}$

The probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.52\\ =0.48\end{array}$

Substitute $p=0.52,q=0.48,k=0,n=10$ in the above formula to find the probability that none of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(0){=}^{10}{C}_0{(}^{0.52}{(}^{0.48}\\ =1(1){(}^{0.48}\\ =0.0006\\ =0.06\%\end{array}$

Hence, the probability that none of the $10$ random students surveyed own a laptop is $\boxed{0.0006or0.06\%}$ .

Substitute $p=0.52,q=0.48,k=1,n=10$ in the above formula to find the probability that exactly one of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(1){=}^{10}{C}_1{(}^{0.52}{(}^{0.48}\\ =10(0.52){(}^{0.48}\\ =0.0007\\ =0.07\%\end{array}$

Hence, the probability that exactly one of the $10$ random students surveyed own a laptop is $\boxed{0.0007or0.07\%}$ .

Substitute $p=0.52,q=0.48,k=2,n=10$ in the above formula to find the probability that exactly $2$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(2){=}^{10}{C}_2{(}^{0.52}{(}^{0.48}\\ =45{(}^{0.52}{(}^{0.48}\\ =0.0343\\ =3.43\%\end{array}$

Hence, the probability that exactly $2$ of the $10$ random students surveyed own a laptop is $\boxed{0.0343or3.43\%}$ .

Substitute $p=0.52,q=0.48,k=3,n=10$ in the above formula to find the probability that exactly $3$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(3){=}^{10}{C}_3{(}^{0.52}{(}^{0.48}\\ =120{(}^{0.52}{(}^{0.48}\\ =0.0991\\ =9.91\%\end{array}$

Hence, the probability that exactly $3$ of the $10$ random students surveyed own a laptop is $\boxed{0.0991or9.91\%}$ .

Substitute $p=0.52,q=0.48,k=4,n=10$ in the above formula to find the probability that exactly $4$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(4){=}^{10}{C}_4{(}^{0.52}{(}^{0.48}\\ =210{(}^{0.52}{(}^{0.48}\\ =0.1878\\ =18.78\%\end{array}$

Hence, the probability that exactly $4$ of the $10$ random students surveyed own a laptop is $\boxed{0.1878or18.78\%}$ .

Substitute $p=0.52,q=0.48,k=5,n=10$ in the above formula to find the probability that exactly $5$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(5){=}^{10}{C}_5{(}^{0.52}{(}^{0.48}\\ =252{(}^{0.52}{(}^{0.48}\\ =0.2441\\ =24.41\%\end{array}$

Hence, the probability that exactly $5$ of the $10$ random students surveyed own a laptop is $\boxed{0.2441or24.41\%}$ .

Substitute $p=0.52,q=0.48,k=6,n=10$ in the above formula to find the probability that exactly $6$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(6){=}^{10}{C}_6{(}^{0.52}{(}^{0.48}\\ =210{(}^{0.52}{(}^{0.48}\\ =0.2204\\ =22.04\%\end{array}$

Hence, the probability that exactly $6$ of the $10$ random students surveyed own a laptop is $\boxed{0.2204or22.04\%}$ .

Substitute $p=0.52,q=0.48,k=7,n=10$ in the above formula to find the probability that exactly $7$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(7){=}^{10}{C}_7{(}^{0.52}{(}^{0.48}\\ =120{(}^{0.52}{(}^{0.48}\\ =0.1364\\ =13.64\%\end{array}$

Hence, the probability that exactly $7$ of the $10$ random students surveyed own a laptop is $\boxed{0.1364or13.64\%}$ .

Substitute $p=0.52,q=0.48,k=8,n=10$ in the above formula to find the probability that exactly $8$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(8){=}^{10}{C}_8{(}^{0.52}{(}^{0.48}\\ =45{(}^{0.52}{(}^{0.48}\\ =0.0554\\ =5.54\%\end{array}$

Hence, the probability that exactly $8$ of the $10$ random students surveyed own a laptop is $\boxed{0.0554or5.54\%}$ .

Substitute $p=0.52,q=0.48,k=9,n=10$ in the above formula to find the probability that exactly $9$ of the $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(9){=}^{10}{C}_9{(}^{0.52}{(}^{0.48}\\ =10{(}^{0.52}{(}^{0.48}\\ =0.0133\\ =1.33\%\end{array}$

Hence, the probability that exactly $9$ of the $10$ random students surveyed own a laptop is $\boxed{0.0133or1.33\%}$ .

Substitute $p=0.52,q=0.48,k=10,n=10$ in the above formula to find the probability that all $10$ random students surveyed own a laptop,

  $\begin{array}{l}p(10){=}^{10}{C}_{10}{(}^{0.52}{(}^{0.48}\\ =1{(}^{0.52}{(}^{0.48}\\ =0.0014\\ =1.14\%\end{array}$

Hence, the probability that all $10$ random students surveyed own a laptop is $\boxed{0.0014or1.14\%}$ .

b.

To determine

What is the probability that at least $8$ of the $10$ students own a laptop?

Answer to Problem 21PPS

  $\boxed{0.0701or7.01\%}$

Explanation of Solution

Given information:

According to a recent survey, $52\%$ of high school students own a laptop. Ten random students are chosen.

What is the probability that at least $8$ of the $10$ students own a laptop?

Calculation:

The number of students out of $10$ random students who own a laptop could be $8,9or10$ .

From part a, it is clear that,

  $\begin{array}{l}P(8)=0.0554\\ P(9)=0.0133\\ P(10)=0.0014\end{array}$

Hence, the probability that at least $8$ of the $10$ students surveyed own a laptop is,

  $\text{P}\left(\text{At least 8}\right)=\text{P}\left(\text{8}\right)+\text{P}\left(\text{9}\right)+\text{P}\left(\text{1}0\right)$

  $\begin{array}{l}=0.0554+0.0133+0.0014\\ =0.0701\\ =7.01\%\end{array}$

Hence, the probability that at least $8$ of the $10$ students surveyed own a laptop is $\boxed{0.0701or7.01\%}$ .

c.

To determine

How many students should you expect to own a laptop?

Answer to Problem 21PPS

  $\boxed{5}$

Explanation of Solution

Given information:

According to a recent survey, $52\%$ of high school students own a laptop. Ten random students are chosen.

How many students should you expect to own a laptop?

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 $n=10,p=0.52$ . Substitute these values in the above formula to find the number of students expected to own a laptop,

  $\begin{array}{l}\mu =np\\ =10(0.52)\\ =5.2\end{array}$

  $\approx 5$

Hence, the number of students who are expected to own a laptop is $\boxed{5}$ .