Chapter 6, Problem 53CE

a.

To determine

Create a probability distribution for the number of travelers who plan their trips within two weeks of departure.

Answer to Problem 53CE

The probability distribution for the number of travelers who plan their trips within two weeks of departure is as follows:

Number of travelers	Probability
0	0.0001
1	0.0019
2	0.0116
3	0.0418
4	0.1020
5	0.1768
6	0.2234
7	0.2075
8	0.1405
9	0.0676
10	0.0220
11	0.0043
12	0.0004

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 travelers who plan their trips within two weeks of departure.

Here, n=12; $\pi =0.52$

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

$\begin{array}{c}P\left(0\right)=C{}_0{\left(0.52\right)}^0{\left(1-0.52\right)}^{\left(12-0\right)}\\ =\frac{12!}{0!\left(12-0\right)!}{\left(0.52\right)}^0{\left(0.48\right)}^{12}\\ =1\times 1\times 0.0001\\ =0.0001\end{array}$

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

Thus, the probability distribution for the number of travelers who plan their trips within two weeks of departure is obtained as follows:

Number of travelers	Probability
0	$C{}_0{\left(0.52\right)}^0{\left(1-0.52\right)}^{\left(12-0\right)}=0.0001$
1	$C{}_1{\left(0.52\right)}^1{\left(1-0.52\right)}^{\left(12-1\right)}=0.0019$
2	$C{}_2{\left(0.52\right)}^2{\left(1-0.52\right)}^{\left(12-2\right)}=0.0116$
3	$C{}_3{\left(0.52\right)}^3{\left(1-0.52\right)}^{\left(12-3\right)}=0.0418$
4	$C{}_4{\left(0.52\right)}^4{\left(1-0.52\right)}^{\left(12-4\right)}=0.1020$
5	$C{}_5{\left(0.52\right)}^5{\left(1-0.52\right)}^{\left(12-5\right)}=0.1768$
6	$C{}_6{\left(0.52\right)}^6{\left(1-0.52\right)}^{\left(12-6\right)}=0.2234$
7	$C{}_7{\left(0.52\right)}^7{\left(1-0.52\right)}^{\left(12-7\right)}=0.2075$
8	$C{}_8{\left(0.52\right)}^8{\left(1-0.52\right)}^{\left(12-8\right)}=0.1405$
9	$C{}_9{\left(0.52\right)}^9{\left(1-0.52\right)}^{\left(12-9\right)}=0.0676$
10	$C{}_{10}{\left(0.52\right)}^{10}{\left(1-0.52\right)}^{\left(12-10\right)}=0.0220$
11	$C{}_{11}{\left(0.52\right)}^{11}{\left(1-0.52\right)}^{\left(12-11\right)}=0.0043$
12	$C{}_{12}{\left(0.52\right)}^{12}{\left(1-0.52\right)}^{\left(12-12\right)}=0.0004$

b.

To determine

Compute the mean and standard deviation for the given distribution.

Answer to Problem 53CE

The mean is 6.24.

The standard deviation is 1.7307.

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:

$\begin{array}{c}\mu =12\times 0.52\\ =6.24\end{array}$

Therefore, the mean of the distribution is 6.24.

The standard deviation of the distribution is calculated as follows:

$\begin{array}{c}\sigma =\sqrt{12\times 0.52\times \left(1-0.52\right)}\\ =\sqrt{2.9952}\\ =1.7307\end{array}$

Therefore, the standard deviation of the distribution is 1.7307.

c.

To determine

Compute the probability of exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure.

Answer to Problem 53CE

The probability of exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure is 0.1775.

Explanation of Solution

The probability of exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure is calculated as follows:

$\begin{array}{c}P\left(x=5\right)=C{}_5{\left(0.52\right)}^5{\left(1-0.52\right)}^{\left(12-5\right)}\\ =\frac{12!}{5!\left(12-5\right)!}\times {0.52}^5\times {0.48}^7\\ =792\times 0.0380\times 0.0059\\ =0.1775\end{array}$

Therefore, the probability of exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure is 0.1775.

d.

To determine

Compute the probability that 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure.

Answer to Problem 53CE

The probability that 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure is 0.3343.

Explanation of Solution

The probability that 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure is calculated as follows:

$\begin{array}{c}P\left(x\le 5\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)\\ =\left\{\begin{array}{c}\left[C{}_0{\left(0.52\right)}^0{\left(1-0.52\right)}^{\left(12-0\right)}\right]+\left[C{}_1{\left(0.52\right)}^1{\left(1-0.52\right)}^{\left(12-1\right)}\right]+\\ \left[C{}_2{\left(0.52\right)}^2{\left(1-0.52\right)}^{\left(12-2\right)}\right]+\left[C{}_3{\left(0.52\right)}^3{\left(1-0.52\right)}^{\left(12-3\right)}\right]+\\ \left[C{}_4{\left(0.52\right)}^4{\left(1-0.52\right)}^{\left(12-4\right)}\right]+\left[C{}_5{\left(0.52\right)}^5{\left(1-0.52\right)}^{\left(12-5\right)}\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\frac{12!}{0!\left(12-0\right)!}\times {0.52}^0\times {0.48}^{12}\right]+\left[\frac{12!}{1!\left(12-1\right)!}\times {0.52}^1\times {0.48}^{11}\right]+\\ \left[\frac{12!}{2!\left(12-2\right)!}\times {0.52}^2\times {0.48}^{10}\right]+\left[\frac{12!}{3!\left(12-3\right)!}\times {0.52}^3\times {0.48}^9\right]+\\ \left[\frac{12!}{4!\left(12-4\right)!}\times {0.52}^4\times {0.48}^8\right]+\left[\frac{12!}{5!\left(12-5\right)!}\times {0.52}^5\times {0.48}^7\right]\end{array}\right\}\\ =\left(0.0002+0.0019+0.0116+0.0418+0.1020+0.1768\right)\\ =0.3343\end{array}$

Therefore, the probability that 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure is 0.3343.