Chapter 14, Problem 14.25E

(a)

To determine

To find: The values of n and p, X has a binomial distribution.

Answer to Problem 14.25E

The value of n and p, if X has a binomial distribution is 5 and 0.25, respectively.

Explanation of Solution

Given info:

Yosemite National park has arranged the 5 high sierra camps $\left(n\right)$ in a 49-mile loop. The random variable X is “the number of years that have been selected to hike the loop”. Also, approximately 25% of the groups are randomly selected to hike the loop.

Justification;

Define the random variable X “the number of years that have been selected to hike the loop.

Also, there are two possible outcomes (groups that hike the loop and do not hike the loop) and the probability of success is the probability that the number of years that have been selected to hike the loop (p) is 0.25 and not selected to hike the loop is 0.75 $\left(=1-0.25\right)$. Thus, X follows the binomial distribution.

Therefore, the number of years that have been selected to hike the loop follows the binomial distribution with sample size $\left(n\right)$ of 5 and the probability $\left(p\right)$ with 0.25.

Thus, the value of n and p, if X has a binomial distribution is 5 and 0.25, respectively.

(b)

To determine

To find: The possible values that X takes.

Answer to Problem 14.25E

The possible values that the random variable X take are 0, 1, 2, 3, 4 and 5.

Explanation of Solution

Since, the national park has arranged only the 5 high sierra camps, the possible values that (X) the number of years that have been selected to hike the loop takes are 0, 1, 2, 3, 4 and 5.

(c)

To determine

To find: The probability value for $X=0$

To find: The probability value for $X=1$

To find: The probability value for $X=2$

To find: The probability value for $X=3$

To find: The probability value for $X=4$

To find: The probability value for $X=5$

To obtain: The probability histogram for the distribution of X

Answer to Problem 14.25E

The probability value with $X=0$ is 0.2373.

The probability value with $X=1$ is 0.3955.

The probability value with $X=2$ is 0.2637.

The probability value with $X=3$ is 0.0879.

The probability value with $X=4$ is 0.0146.

The probability value with $X=5$ is 0.0010.

The probability histogram for the distribution of X is given below:

Explanation of Solution

Given info:

The probability value with $X=0$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=5$, p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}P\left(X=0\right)=\frac{5!}{0!\left(5-0\right)!}{\left(0.25\right)}^0{\left(0.75\right)}^{5-0}\\ =1\times 1\times 0.2373\\ =0.2373\end{array}$

Thus, the probability value with $X=0$ is 0.2373.

The probability value with $X=1$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=5$, p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}P\left(X=1\right)=\frac{5!}{1!\left(5-1\right)!}{\left(0.25\right)}^1{\left(0.75\right)}^{5-1}\\ =5\times 0.25\times \mathrm{0.0.3164}\\ =0.3955\end{array}$

Thus, the probability value with $X=1$ is 0.3955.

The probability value with $X=2$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=5$,p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}P\left(X=2\right)=\frac{5!}{2!\left(5-2\right)!}{\left(0.25\right)}^2{\left(0.75\right)}^{5-2}\\ =10\times 0.0625\times 0.4219\\ =0.2637\end{array}$

Thus, the probability value with $X=2$ is 0.2637.

The probability value with $X=3$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=5$, p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}P\left(X=3\right)=\frac{5!}{3!\left(5-3\right)!}{\left(0.25\right)}^3{\left(0.75\right)}^{5-3}\\ =10\times 0.0157\times 0.5625\\ =0.0879\end{array}$

Thus, the probability value with $X=3$ is 0.0879

The probability value with $X=4$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=5$, p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}P\left(X=4\right)=\frac{5!}{4!\left(5-4\right)!}{\left(0.25\right)}^4{\left(0.75\right)}^{5-4}\\ =5\times 0.0039\times 0.75\\ =0.0146\end{array}$

Thus, the probability value with $X=4$ is 0.0146.

The probability value with $X=5$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=5$, p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}P\left(X=5\right)=\frac{5!}{5!\left(5-5\right)!}{\left(0.25\right)}^5{\left(0.75\right)}^{5-5}\\ =1\times 0.0010\times 1\\ =0.0010\end{array}$

Thus, the probability value with $X=5$ is 0.0010.

Software procedure:

Step-by-step procedure to obtain the probability histogram for the distribution of X is:

• Choose Graph > Bar Chart.
• From Bars represent, choose Values from a table.
• Under One column of values, choose Simple. Click OK.
• In Graph variables, enter the column of X as
• In Row labels, enter the column of P(X)
• Click OK.

Justification:

In the probability histogram, the values of X are taken along the horizontal axis, the probability values along the vertical axis, and the corresponding probability is represented by the height of the bars.

(d)

To determine

To find: The mean and the standard deviation of the distribution.

Answer to Problem 14.25E

The mean and the standard deviation of the distribution is 1.25 years and 0.9682 years.

Explanation of Solution

Calculation:

Mean:

The mean is calculated by using the formula:

$\mu =np$

Substitute n as 5 and p as 0.25.

$\begin{array}{c}\mu =\left(5\right)\left(0.25\right)\\ =1.25\end{array}$

Thus, the mean of distribution is 1.25 years.

Standard deviation:

The standard deviation is calculated by using the formula:

$\sigma =\sqrt{np\left(1-p\right)}$

Substitute n as 5, p as 0.25 and q as $0.75\left(=1-0.25\right)$

$\begin{array}{c}\sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{\left(5\right)\left(0.25\right)\left(0.75\right)}\\ =\sqrt{0.9375}\\ =0.9682\end{array}$

Thus, the standard deviation of distribution is 1.2962 years.

From the part (c), by using the probability histogram for the distribution of X, the location of mean with 1.25 is marked on the histogram as follows: