Chapter 8.1, Problem 13E

(a)

To determine

Graph for density function.

Explanation of Solution

Suppose the given density function is $\text{F}\left(\text{X}\right)=\{\begin{array}{l}\frac{\text{X}}{25}\text{; }\left(\text{0}<\text{x}<5\right)\\ \frac{\text{10}-\text{X}}{25}\text{; }\left(\text{5}<\text{x}<10\right)\end{array}$.

The density function can be represented as follows:

(b)

To determine

Probability of X that lies between 1 and 3.

Explanation of Solution

The probability of X that lies between 1 and 3 can be obtained as follows:

$\begin{array}{c}\text{P}\left(1<\text{X}<3\right)=\text{P}\left(\text{X}<3\right)-\text{P}\left(\text{X}<1\right)\\ =\frac{1}{2}\times \frac{3}{25}\times \left(3-0\right)-\frac{1}{2}\times \frac{1}{25}\times \left(1-0\right)\\ =0.18-0.2\\ =0.16\end{array}$

(c)

To determine

Probability of X that lies between 4 and 8.

Explanation of Solution

The probability can be estimated as follows:

$\begin{array}{c}\text{P}\left(4<\text{X}<8\right)=\text{P}\left(4<\text{X}<5\right)+\text{P}\left(5<\text{X}<8\right)\mathrm{......}(1)\\ \text{P}\left(4<\text{X}<5\right)=\text{P}\left(\text{X}<5\right)-\text{P}\left(\text{X}<4\right)\\ =\frac{1}{2}\times \frac{5}{25}\times \left(5-0\right)-\frac{1}{2}\times \frac{4}{25}\times \left(4-0\right)\\ =0.5-0.32\\ =0.18\\ \text{P}\left(5<\text{X}<8\right)=\text{P}\left(\text{X}>5\right)-\text{P}\left(\text{X}>8\right)\\ =\frac{1}{2}\times \frac{5}{25}\times \left(10-5\right)-\frac{1}{2}\times \frac{2}{25}\times \left(10-8\right)\\ =0.5-0.8\\ =0.42\end{array}$

Substitute the values in the above Equation (1).

$\begin{array}{c}\text{P}\left(4<\text{X}<8\right)=\text{P}\left(4<\text{X}<5\right)+\text{P}\left(5<\text{X}<8\right)\\ =0.18+0.42\\ =0.60\end{array}$

(d)

To determine

Estimate the probability of X less than 7.

Explanation of Solution

The probability can be estimated as follows:

$\begin{array}{c}\text{P}\left(\text{X}<7\right)=1-\text{P}\left(\text{X}>7\right)\\ \text{P}\left(\text{X}>7\right)=\frac{1}{2}\times \frac{3}{25}\times \left(10-7\right)\\ =0.18\\ \text{P}\left(\text{X}<7\right)=1-0.18\\ =0.82\end{array}$

(e)

To determine

Estimate the probability that X is greater than 3.

Explanation of Solution

The probability can be estimated as follows:

$\begin{array}{c}\text{P}\left(\text{X}>3\right)=1-\text{P}\left(\text{X}<3\right)\\ \text{P}\left(\text{X}<3\right)=\frac{1}{2}\times \frac{3}{25}\times \left(13-0\right)\\ =0.18\\ \text{P}\left(\text{X}>3\right)=1-0.18\\ =0.82\end{array}$