Chapter 6, Problem 52CE

a.

To determine

Create a probability distribution for the number of students in the sample who watch soap operas.

Answer to Problem 52CE

The probability distribution for the number of students in the sample who watch soap operas is as follows:

Number of students	Probability
0	0.0025
1	0.0207
2	0.0763
3	0.1665
4	0.2384
5	0.2340
6	0.1596
7	0.0746
8	0.0229
9	0.0042
10	0.0003

Explanation of Solution

The formula to find the binomial probability is as follows:

$\begin{array}{l}P\left(X\right)=C{}_x{\pi }^x{\left(1-\pi \right)}^{\left(n-x\right)}\\ \text{where, }C\text{ is the combination}\text{.}\\ n\text{ is the number of trials}\text{.}\\ X\text{ is the random variable}\text{.}\\ \pi \text{ is the probability of success}\text{.}\end{array}$

Let x denotes the students who watch soap operas.

Here, n=10; $\pi =0.45$

For the number of students $x=0$, the probability is calculated as follows:

$\begin{array}{c}P\left(0\right)=C{}_0{\left(0.45\right)}^0{\left(1-0.45\right)}^{\left(10-0\right)}\\ =\frac{10!}{0!\left(10-0\right)!}{\left(0.45\right)}^0{\left(0.55\right)}^{10}\\ =1\times 1\times 0.0025\\ =0.0025\end{array}$

Similarly the probability value for the remaining x values is calculated.

Thus, the probability distribution for the number of students in the sample who watch soap operas is obtained as follows:

Number of students	Probability
0	$C{}_0{\left(0.45\right)}^0{\left(1-0.45\right)}^{\left(10-0\right)}=0.0025$
1	$C{}_1{\left(0.45\right)}^1{\left(1-0.45\right)}^{\left(10-1\right)}=0.0207$
2	$C{}_2{\left(0.45\right)}^2{\left(1-0.45\right)}^{\left(10-2\right)}=0.0763$
3	$C{}_3{\left(0.45\right)}^3{\left(1-0.45\right)}^{\left(10-3\right)}=0.1665$
4	$C{}_4{\left(0.45\right)}^4{\left(1-0.45\right)}^{\left(10-4\right)}=0.2384$
5	$C{}_5{\left(0.45\right)}^5{\left(1-0.45\right)}^{\left(10-5\right)}=0.2340$
6	$C{}_6{\left(0.45\right)}^6{\left(1-0.45\right)}^{\left(10-6\right)}=0.1596$
7	$C{}_7{\left(0.45\right)}^7{\left(1-0.45\right)}^{\left(10-7\right)}=0.0746$
8	$C{}_8{\left(0.45\right)}^8{\left(1-0.45\right)}^{\left(10-8\right)}=0.0229$
9	$C{}_9{\left(0.45\right)}^9{\left(1-0.45\right)}^{\left(10-9\right)}=0.0042$
10	$C{}_{10}{\left(0.45\right)}^{10}{\left(1-0.45\right)}^{\left(10-10\right)}=0.0003$

b.

To determine

Compute the mean and standard deviation of this distribution.

Answer to Problem 52CE

The mean is 4.5.

The standard deviation is 1.5732.

Explanation of Solution

The mean and standard deviation of the binomial distribution are as follows:

$\begin{array}{l}\mu =n\pi \\ \sigma =\sqrt{n\pi \left(1-\pi \right)}\end{array}$

The mean of the distribution is calculated as follows:

$\begin{array}{c}\mu =10\times 0.45\\ =4.5\end{array}$

Therefore, the mean of the distribution is 4.5.

The standard deviation of the distribution is calculated as follows:

$\begin{array}{c}\sigma =\sqrt{10\times 0.45\times \left(1-0.45\right)}\\ =\sqrt{2.475}\\ =1.5732\end{array}$

Therefore, the standard deviation of the distribution is 1.5732.

c.

To determine

Compute the probability of finding exactly four students who watch soap operas.

Answer to Problem 52CE

The probability of finding exactly four students who watch soap operas is 0.2384.

Explanation of Solution

The probability of finding exactly four students who watch soap operas is calculated as follows:

$\begin{array}{c}P\left(x=4\right)=C{}_4{\left(0.45\right)}^4{\left(1-0.45\right)}^{\left(10-4\right)}\\ =\frac{10!}{4!\left(10-4\right)!}\times {0.45}^4\times {0.55}^6\\ =210\times 0.0410\times 0.0277\\ =0.2384\end{array}$

Therefore, the probability of finding exactly four students who watch soap operas is 0.2384.

d.

To determine

Compute the probability that less than half of the students selected watch soap operas.

Answer to Problem 52CE

The probability that less than half of the students selected watch soap operas is 0.5044.

Explanation of Solution

The probability that less than half of the students selected watch soap operas is calculated as follows:

$\begin{array}{c}P\left(x<5\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)\\ =\left\{\begin{array}{c}\left[C{}_0{\left(0.45\right)}^0{\left(1-0.45\right)}^{\left(10-0\right)}\right]+\left[C{}_1{\left(0.45\right)}^1{\left(1-0.45\right)}^{\left(10-1\right)}\right]+\\ \left[C{}_2{\left(0.45\right)}^2{\left(1-0.45\right)}^{\left(10-2\right)}\right]+\left[C{}_3{\left(0.45\right)}^3{\left(1-0.45\right)}^{\left(10-3\right)}\right]+\\ \left[C{}_4{\left(0.45\right)}^4{\left(1-0.45\right)}^{\left(10-4\right)}\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\frac{10!}{0!\left(10-0\right)!}\times {0.45}^0\times {0.55}^{10}\right]+\left[\frac{10!}{1!\left(10-1\right)!}\times {0.45}^1\times {0.55}^9\right]+\\ \left[\frac{10!}{2!\left(10-2\right)!}\times {0.45}^2\times {0.55}^8\right]+\left[\frac{10!}{3!\left(10-3\right)!}\times {0.45}^3\times {0.55}^7\right]+\\ \left[\frac{10!}{4!\left(10-4\right)!}\times {0.45}^4\times {0.55}^6\right]\end{array}\right\}\\ =\left(0.0025+0.0207+0.0763+0.1665+0.2384\right)\\ =0.5044\end{array}$

Therefore, the probability that less than half of the students selected watch soap operas is 0.5044.