Answer to Question #178061 in Statistics and Probability for Asoyine Anthony

Question

 A charity group raises funds by collecting waste paper. A skip-full will contain an 

amount, X, of other materials such as plastic bags and rubber bands. X may be regarded 

as a random variable with probability function

𝑓(𝑥) = {𝑘(𝑥 − 1)(4 − 𝑥) , 1 < 𝑥 < 4

0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

 (All numerical values in this question are in units of 100kg,)

i )Show that 𝑘 =1/2.

 ii)Find the mean and the standard deviation of X.

Answer 1

i) k we will find from the properties of the density function $\int\limits_{ - \infty }^{ + \infty } {f(x)dx = 1}$ . Then

$\int\limits_1^4 {k(x - 1)(4 - x)dx = 1} \Rightarrow \int\limits_1^4 {\left( {5x - 4 - {x^2}} \right)dx = 1} \Rightarrow k\left( {\left. {\frac{{5{x^2}}}{2}} \right|_1^4 - 4\left. x \right|_1^4 - \left. {\frac{{{x^3}}}{3}} \right|_1^4} \right) = 1\Rightarrow$

${k\left( {\frac{{80 - 5}}{2} - 4(4 - 1) - \frac{{64 - 1}}{3}} \right) = 1 \Rightarrow k\left( {\frac{{75}}{2} - 12 - 21} \right) = 1 \Rightarrow }{\frac{9}{2}}{k = 1 \Rightarrow k = \frac{2}{9} \ne \frac{1}{2}}$

Therefore, k cannot be equal to 1/2 and this statement is not true.

ii) Find the mean

$M(x) = \int\limits_{ - \infty }^{ + \infty } {xf(x)dx} = \frac{2}{9}\int\limits_1^4 {x\left( {x - 1} \right)\left( {4 - x} \right)dx = } \frac{2}{9}\int\limits_1^4 {\left( { - {x^3} + 5{x^2} - 4x} \right)} dx = \frac{2}{9}\left( { - \left. {\frac{{{x^4}}}{4}} \right|_1^4 + \left. {\frac{{5{x^3}}}{3}} \right|_1^4 - 2\left. {{x^2}} \right|_1^4} \right) = \frac{2}{9}\left( { - \frac{{256 - 1}}{4} + \frac{{320 - 5}}{3} - 2(16 - 1)} \right) = \frac{{5}}{2}$

Find the variance:

$D(x) = \int\limits_{ - \infty }^{ + \infty } {{x^2}f(x)dx} - {M^2}(x) = \frac{2}{9}\int\limits_1^4 {\left( { - {x^4} + 5{x^3} - 4{x^2}} \right)} dx - {\left( {\frac{5}{2}} \right)^2} = \frac{2}{9}\left( { - \left. {\frac{{{x^5}}}{5}} \right|_1^4 + \left. {\frac{{5{x^4}}}{4}} \right|_1^4 - \left. {\frac{{4{x^3}}}{3}} \right|_1^4} \right) - \frac{{25}}{4} = \frac{2}{9}\left( { - \frac{{1024 - 1}}{5} + \frac{{1280 - 5}}{4} - \frac{{256 - 4}}{3}} \right) - \frac{{25}}{4} = \frac{9}{{20}}$

Then the standard deviation is

$\sigma \left( x \right) = \sqrt {D(x)} = \sqrt {\frac{9}{{20}}} = \frac{3}{{2\sqrt 5 }}$

Answer: $M(x) = \frac{5}{2},\,\sigma \left( x \right) = \frac{3}{{2\sqrt 5 }}$