We know that "X \\sim U(20,40)" with probability density
"{p_X}(x)=\\begin{cases} 0 & x \\notin [a,b] \\\\ \\frac{1}{{b - a}} & x \\in [a,b]\\end{cases}" Let's find the probability density as
"{F_X}(x) = P(X < x) = \\int\\limits_{ - \\infty }^x {{p_X}(\\xi )d\\xi }"At the interval "x \\in ( - \\infty ,a]" we get "{F_X}(x) = \\int\\limits_{ - \\infty }^x {0d\\xi } = 0"
At the interval "x \\in [a,b]" we get "{F_X}(x) = \\int\\limits_a^x {\\frac{1}{{b - a}}d\\xi } = \\left. {\\frac{\\xi }{{b - a}}} \\right|_a^x = \\frac{{x - a}}{{b - a}}"
At the interval "x \\in [b, + \\infty )" we get "{F_X}(x) = \\frac{{b - a}}{{b - a}} + \\int\\limits_b^{ + \\infty } {0d\\xi } = 1"
Thus we get
"{F_X}(x)=\\begin{cases} 0 & x <a \\\\ \\frac{{x - a}}{{b - a}} & a \\leqslant x \\leqslant b \\\\ 1 & x > b \\end{cases}" a) We asked to find "P(X < 35)"
"P(X < 35) = {F_X}(35) = \\frac{{35 - 20}}{{40 - 20}} = \\frac{3}{4}" b) We asked to find "P(23 < X < 32)"
"P(23 < X < 32) = {F_X}(32) - {F_X}(23) = \\frac{{32 - 20}}{{40 - 20}} - \\frac{{23 - 20}}{{40 - 20}} = \\frac{9}{{20}}" c) We asked to find the expected value and standard deviation. Let's calculate the expected value using it's definition
"\\mathbb{E}X = \\int\\limits_{ - \\infty }^{ + \\infty } {x{p_X}(x)dx} = \\int\\limits_a^b {\\frac{x}{{b - a}}dx} = \\left. {\\frac{{{x^2}}}{{2(b - a)}}} \\right|_a^b = \\frac{{{b^2} - {a^2}}}{{2(b - a)}} = \\frac{{a + b}}{2}"
Thus in our case
"\\mathbb{E}X = \\frac{{a + b}}{2} = \\frac{{20 + 40}}{2} = 30" To find the standart deviatin we need to find the variance at the first. To find the variance let's find the second moment
"\\mathbb{E}({X^2}) = \\int\\limits_{ - \\infty }^{ + \\infty } {{x^2}{p_X}(x)dx} = \\int\\limits_a^b {\\frac{{{x^2}}}{{b - a}}dx} = \\left. {\\frac{{{x^3}}}{{3(b - a)}}} \\right|_a^b = \\frac{{{a^2} + ab + {b^2}}}{3}"
Now we can calculate the variance
"\\operatorname{Var} (X) = \\mathbb{E}({X^2}) - {\\mathbb{E}^2}(X) = \\frac{{{a^2} + ab + {b^2}}}{3} - \\frac{{{{(a + b)}^2}}}{{{2^2}}} = \\frac{1}{{12}}{(b - a)^2}" And the standard deviations is
"\\sigma (X) = \\sqrt {\\frac{1}{{12}}{{(a - b)}^2}} = \\frac{{b - a}}{{2\\sqrt 3 }}" In our case
"\\sigma (X) = \\frac{{40 - 20}}{{2\\sqrt 3 }} = \\frac{{10}}{{\\sqrt 3 }}"
Comments
Dear Azlan Ikram, thank you for correcting us.
Its (x-a)/(b-a) for a
Leave a comment