Chapter 6, Problem 37RE

a.

To determine

To explain: why the given function is a probability density function.

Explanation of Solution

Given information: Given function is $\begin{array}{l}f(x)=\frac{\pi }{20}\sin (\frac{\pi x}{10}),\text{if}0\le x\le 10.\\ f(x)=0,\text{if}x<0\text{or}x>10.\end{array}$

Calculation:

In general, the probability density function of a random variable x satisfies the condition x $\ge 0$ for all x .

The given function is always greater or equal to zero so this condition is satisfied:

Next thing need to prove is that for given function f(x) the following is true:

  ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f(x)dx}=1$

This is fairly easy to prove.

  $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f(x)dx}={\displaystyle \underset{-\infty }{\overset{0}{\int }}(0)dx}+{\displaystyle \underset{0}{\overset{10}{\int }}\frac{\pi }{20}}\sin (\frac{\pi x}{10})dx+{\displaystyle \underset{10}{\overset{\infty }{\int }}(0)dx}\\ =0+{\displaystyle \underset{0}{\overset{10}{\int }}\frac{\pi }{20}}\sin (\frac{\pi x}{10})d+0\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{10}{\int }}\frac{\pi }{10}}\sin (\frac{\pi x}{10})\\ =\frac{-1}{2}{\displaystyle \underset{0}{\overset{10}{\int }}(-\frac{\pi }{10}}\sin (\frac{\pi x}{10}))\\ =\frac{-1}{2}\cos (\frac{\pi x}{10})|\begin{array}{c}10\\ 0\end{array}\\ =-\frac{1}{2}[\cos (\frac{\pi (10)}{10})-\cos (\frac{\pi (0)}{10})]\\ =-\frac{1}{2}[\cos \pi -\cos (0)]\\ =-\frac{1}{2}[-1-1]\\ =1\end{array}$

Hence proved, f(x) is a probability density function.

b.

To determine

To find: $P(X<4).$

Answer to Problem 37RE

  $P(X<4)=0.3455$ .

Explanation of Solution

Given information: Given function is $\begin{array}{l}f(x)=\frac{\pi }{20}\sin (\frac{\pi x}{10}),\text{if}0\le x\le 10.\\ f(x)=0,\text{if}x<0\text{or}x>10.\end{array}$

Calculation:

Since 1 is the total probability, so:

  $\begin{array}{c}P(X<4)=1-P(x\ge 4)\\ ={\displaystyle \underset{4}{\overset{10}{\int }}f(x)dx}\\ ={\displaystyle \underset{0}{\overset{10}{\int }}\frac{\pi }{20}}\sin (\frac{\pi x}{10})\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{10}{\int }}\frac{\pi }{10}}\sin (\frac{\pi x}{10})\\ =1-\frac{1}{2}{\displaystyle \underset{4}{\overset{10}{\int }}\frac{\pi }{10}}\sin (\frac{\pi x}{10})\\ =1+\frac{1}{2}{\displaystyle \underset{4}{\overset{10}{\int }}(-\frac{\pi }{10}}\sin (\frac{\pi x}{10}))\\ =1+\frac{1}{2}\cos (\frac{\pi x}{10})|\begin{array}{c}10\\ 4\end{array}\\ =1+\frac{1}{2}[\cos (\frac{\pi (10)}{10})-\cos (\frac{\pi (4)}{10})]\\ =1+\frac{1}{2}[-1-0.309]\\ =1-\frac{1}{2}(1.309)\\ =1-0.6545\\ =0.3455\\ P(X<4)=\mathrm{0.3455.}\end{array}$

c.

To determine

To calculate: the mean and find is the value would expect.

Answer to Problem 37RE

Mean value is 5.

Explanation of Solution

Given information: Given function is $\begin{array}{l}f(x)=\frac{\pi }{20}\sin (\frac{\pi x}{10}),\text{if}0\le x\le 10.\\ f(x)=0,\text{if}x<0\text{or}x>10.\end{array}$

Formula used:

The mean of the any probability density function f(x) is given by;

  $\mu ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf(x)dx}$ .

Calculation:

Since the function is zero outside the interval[0,10], so;

  $\begin{array}{c}\mu ={\displaystyle \underset{0}{\overset{10}{\int }}xf(x)dx}=\frac{\pi }{20}{\displaystyle \underset{}{\overset{}{\int }}x\sin (\frac{\pi x}{10}})dx.\\ \mu =\frac{1}{2}{\displaystyle \underset{0}{\overset{10}{\int }}\frac{10}{\pi }}.[\frac{\pi }{10}x]\sin (\frac{\pi x}{10})[\frac{\pi }{10}dx]\\ \text{Substitute}\frac{\pi x}{10}=u\text{and}\frac{\pi }{10}dx=du.\\ \text{Limits}\text{of}\text{integration}\text{will}\text{change}\text{from}{\displaystyle \underset{0}{\overset{10}{\int }}to{\displaystyle \underset{0}{\overset{\pi }{\int }}}}\\ \mu =\frac{5}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}u\sin udu}\\ =\frac{5}{\pi }[u{\displaystyle \int \sin u-{\displaystyle \int [\frac{du}{du}}}{\displaystyle \int \sin udu}]du]\begin{array}{c}\pi \\ 0\end{array}\\ =\frac{5}{\pi }[-u\cos u+{\displaystyle \int \cos udu]}\begin{array}{c}\pi \\ 0\end{array}\\ =\frac{5}{\pi }[-u\cos u+\sin u]\begin{array}{c}\pi \\ 0\end{array}\\ =\frac{5}{\pi }.\pi =5\\ \mu =5.\end{array}$