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

Question #178061

 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.


1
Expert's answer
2021-04-15T07:39:45-0400

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 }}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS