Chapter 3.1, Problem 3.8E

(a)

To determine

To explain the reason due to which the height of the curve is equal to 1.

Explanation of Solution

Given:

  $x~\text{Uniform(0, 2)}$

Graph:

  

The height of the given graph is 0.5, because it is equally distributed for the range 0 − 1 as 0.5 and 1 − 2 as 0.5 and the area under the curve is equal to 1. The height of the uniform distribution defines the proportion of the area per 1 unit.

(b)

To determine

To calculate the proportion that falls between 1 and 1.4.

Answer to Problem 3.8E

0.7

Explanation of Solution

Calculation:

  $\begin{array}{c}\text{Proportion }\Rightarrow {\displaystyle {\int }_0^{1.4}\frac{1}{2-0}dx}\\ \Rightarrow \frac{1}{2}{\left[x\right]}_0^{1.4}\\ \Rightarrow \frac{1}{2}\left[1.4-0\right]\\ \Rightarrow 0.7\end{array}$

(c)

To determine

To calculate the median and quartiles for given distribution.

Answer to Problem 3.8E

First quartile = 0.5

Median = 1

Third quartile = 1.5

Explanation of Solution

Calculation:

  $\begin{array}{c}\text{First quartile =}{\displaystyle {\int }_0^x\frac{1}{2-0}dx}=0.25\\ \Rightarrow \frac{1}{2}{\left[x\right]}_0^x=0.25\\ \Rightarrow \frac{1}{2}\left[x-0\right]=0.25\\ \Rightarrow x=0.5\end{array}$

  $\begin{array}{c}\text{Median =}{\displaystyle {\int }_0^x\frac{1}{2-0}dx}=0.50\\ \Rightarrow \frac{1}{2}{\left[x\right]}_0^x=0.50\\ \Rightarrow \frac{1}{2}\left[x-0\right]=0.50\\ \Rightarrow x=1\end{array}$

  $\begin{array}{c}\text{Third quartile =}{\displaystyle {\int }_0^x\frac{1}{2-0}dx}=0.75\\ \Rightarrow \frac{1}{2}{\left[x\right]}_0^x=0.75\\ \Rightarrow \frac{1}{2}\left[x-0\right]=0.75\\ \Rightarrow x=1.50\end{array}$