Answer to Question #169785 in Statistics and Probability for Amedeka Israel Yao

Question #169785

The continuous random variable x has the probability density function,

𝑓(π‘₯) =

{

1/ 2(π‘₯ - 2), 2 ≀ π‘₯ ≀ 3

π‘Ž, 3 < π‘₯ ≀ 5

2 - 𝑏π‘₯, 5 < π‘₯ ≀ 6

0, π‘’π‘™π‘ π‘’π‘€β„Žπ‘’π‘Ÿπ‘’

i. Find the values of a and b, and sketch the graph of f(x).

ii. Find the cumulative distribution function, f(x) and sketch it.


1
Expert's answer
2021-03-09T05:31:22-0500


Find the distribution function "F(x) = \\int\\limits_{ - \\infty }^x {f(x)dx}".

If "x < 2" then "F(x) = \\int\\limits_{ - \\infty }^x {0dx} = 0"

If "x \\le 3" then "F(x) = \\int\\limits_{ - \\infty }^2 {0dx} + \\int\\limits_2^x {\\frac{1}{2}(x - 2)dx} = \\left. {\\frac{{{x^2}}}{4}} \\right|_2^x - \\left. x \\right|_2^x = \\frac{{{x^2} - 4}}{4} - x + 2 = \\frac{{{x^2}}}{4} - x + 1"

If "x \\le 5" then "F(x) = \\int\\limits_{ - \\infty }^2 {0dx} + \\int\\limits_2^3 {\\frac{1}{2}(x - 2)dx} + \\int\\limits_3^x {adx} = \\left. {\\frac{{{x^2}}}{4}} \\right|_2^3 - \\left. x \\right|_2^3 + a\\left. x \\right|_3^x = \\frac{{9 - 4}}{4} - 3 + 2 + a(x - 3) = \\frac{5}{4} - 1 + ax - 3a = ax - 3a + \\frac{1}{4}"

If "x \\le 6" then

"F(x) = \\int\\limits_{ - \\infty }^2 {0dx} + \\int\\limits_2^3 {\\frac{1}{2}(x - 2)dx} + \\int\\limits_3^5 {adx} + \\int\\limits_5^x {\\left( {2 - bx} \\right)} = \\left. {\\frac{{{x^2}}}{4}} \\right|_2^3 - \\left. x \\right|_2^3 + a\\left. x \\right|_3^5 + 2\\left. x \\right|_5^x - \\left. {\\frac{{b{x^2}}}{2}} \\right|_5^x = \\frac{{9 - 4}}{4} - 3 + 2 + 2a + 2(x - 5) - \\frac{b}{2}({x^2} - 25) = \\frac{5}{4} - 1 + 2a + 2x - 10 - \\frac{{b{x^2}}}{2} + \\frac{{25b}}{2} = - \\frac{{b{x^2}}}{2} + 2x + 2a + \\frac{{25b}}{2} - \\frac{{39}}{4}"

If "x > 6" then "F(x) = \\int\\limits_{ - \\infty }^2 {0dx} + \\int\\limits_2^3 {\\frac{1}{2}(x - 2)dx} + \\int\\limits_3^5 {adx} + \\int\\limits_5^6 {\\left( {2 - bx} \\right)} = \\left. {\\frac{{{x^2}}}{4}} \\right|_2^3 - \\left. x \\right|_2^3 + a\\left. x \\right|_3^5 + 2\\left. x \\right|_5^6 - \\left. {\\frac{{b{x^2}}}{2}} \\right|_5^6 = \\frac{{9 - 4}}{4} - 3 + 2 + 2a + 2(6 - 5) - \\frac{b}{2}(36 - 25) = \\frac{5}{4} - 1 + 2a + 2 - \\frac{{11b}}{2} = \\frac{9}{4} + 2a - \\frac{{11b}}{2}"

Then "F(x) = \\left\\{ {\\begin{matrix}\n{0,\\,\\,x < 2}\\\\\n{\\frac{{{x^2}}}{4} - x + 1,\\,\\,2 \\le x \\le 3}\\\\\n{ax - 3a + \\frac{1}{4},\\,\\,3 < x \\le 5}\\\\\n{ - \\frac{{b{x^2}}}{2} + 2x + 2a + \\frac{{25b}}{2} - \\frac{{39}}{4},\\,\\,5 < x \\le 6}\\\\\n{\\frac{9}{4} + 2a - \\frac{{11b}}{2},\\,\\,x > 6}\n\\end{matrix}} \\right."

By the properties of the distribution function

"\\mathop {\\lim }\\limits_{x \\to 3 - 0} F(x) = \\mathop {\\lim }\\limits_{x \\to 3 + 0} F(x) \\Rightarrow \\frac{{{3^2}}}{4} - 3 + 1 = 3a - 3a + \\frac{1}{4}"

"\\mathop {\\lim }\\limits_{x \\to 5 - 0} F(x) = \\mathop {\\lim }\\limits_{x \\to 5 + 0} F(x) \\Rightarrow 5a - 3a + \\frac{1}{4} = - \\frac{{25b}}{2} + 10 + 2a + \\frac{{25b}}{2} - \\frac{{39}}{4}"

"\\mathop {\\lim }\\limits_{x \\to 6 - 0} F(x) = \\mathop {\\lim }\\limits_{x \\to 6 + 0} F(x) \\Rightarrow - \\frac{{36b}}{2} + 12 + 2a + \\frac{{25b}}{2} - \\frac{{39}}{4} = \\frac{9}{4} + 2a - \\frac{{11b}}{2}"

"F(\\infty ) = 1 \\Rightarrow \\frac{9}{4} + 2a - \\frac{{11b}}{2} = 1"

We have the system of equations:

"\\left\\{ \\begin{array}{l}\n\\frac{{{3^2}}}{4} - 3 + 1 = 3a - 3a + \\frac{1}{4}\\\\\n5a - 3a + \\frac{1}{4} = - \\frac{{25b}}{2} + 10 + 2a + \\frac{{25b}}{2} - \\frac{{39}}{4}\\\\\n - \\frac{{36b}}{2} + 12 + 2a + \\frac{{25b}}{2} - \\frac{{39}}{4} = \\frac{9}{4} + 2a - \\frac{{11b}}{2}\\\\\n\\frac{9}{4} + 2a - \\frac{{11b}}{2} = 1\n\\end{array} \\right."


"\\left\\{ \\begin{array}{l}\n\\frac{1}{4} = \\frac{1}{4}\\\\\n2a + \\frac{1}{4} = 2a + \\frac{1}{4}\\\\\n - \\frac{{11b}}{2} + \\frac{9}{4} + 2a = \\frac{9}{4} + 2a - \\frac{{11b}}{2}\\\\\n2a - \\frac{{11b}}{2} = - \\frac{5}{4}\n\\end{array} \\right."


"2a - \\frac{{11b}}{2} = - \\frac{5}{4} \\Rightarrow 8a - 22b = - 5 \\Rightarrow a = \\frac{{ - 5 + 22b}}{8}"

This equation has an infinite number of solutions. Let's find any particular solution.

Let "b = \\frac{5}{{22}} \\Rightarrow a = 0"

Then

"f(x) = \\left\\{ {\\begin{matrix}\n{\\frac{1}{2}(x - 2),\\,\\,2 \\le x \\le 3}\\\\\n{0,\\,\\,3 < x \\le 5}\\\\\n{2 - \\frac{5}{{22}}x,\\,\\,5 < x \\le 6}\\\\\n{0,\\,\\,otherwise}\n\\end{matrix}} \\right."

"F(x) = \\left\\{ {\\begin{matrix}\n{0,\\,\\,x < 2}\\\\\n{\\frac{{{x^2}}}{4} - x + 1,\\,\\,2 \\le x \\le 3}\\\\\n{0 - 0 + \\frac{1}{4},\\,\\,3 < x \\le 5}\\\\\n{ - \\frac{{5{x^2}}}{{44}} + 2x + \\frac{{125}}{{44}} - \\frac{{39}}{4},\\,\\,5 < x \\le 6}\\\\\n{\\frac{9}{4} - \\frac{{55}}{{44}},\\,\\,x > 6}\n\\end{matrix}} \\right.= \\left\\{ {\\begin{matrix}\n{0,\\,\\,x < 2}\\\\\n{\\frac{{{x^2}}}{4} - x + 1,\\,\\,2 \\le x \\le 3}\\\\\n{0 - 0 + \\frac{1}{4},\\,\\,3 < x \\le 5}\\\\\n{ - \\frac{{5{x^2}}}{{44}} + 2x - \\frac{{76}}{{11}},\\,\\,\\,5 < x \\le 6}\\\\\n{1,\\,\\,x > 6}\n\\end{matrix}} \\right."

sketch the graph of f(x):



sketch the graph of F(x):


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