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Question Number 192377 by Spillover last updated on 16/May/23
The probability density function f(x)   of a variable x is given by   f(x)= { ((kxsin πx     0≤x≤1)),((0 for all value of x)) :}  Show that k=π  and deduce that mean   and the variance of the distribution are  (1−(4/π^2 )) and (2/π^2 )(1−(8/π^2 ))
Theprobabilitydensityfunctionf(x)ofavariablexisgivenbyf(x)={kxsinπx0x10forallvalueofxShowthatk=πanddeducethatmeanandthevarianceofthedistributionare(14π2)and2π2(18π2)
Answered by mehdee42 last updated on 16/May/23
∫_0 ^1 kxsinπxdx=1⇒−(k/π)xcosπx+(k/π^2 )sinπx]_0 ^1 =1  ⇒k=π ✓  E(x)=∫_0 ^1 πx^2 sinπxdx=π(−(x^2 /π)cosπx+((2x)/π^2 )sinπx+(2/π^2 )cosπx]_0 ^1   π((1/π)−(4/π^3 ))=1−(4/π^2 )  ✓  E(x^2 )=∫_0 ^1  x^3  sinπxdx  =π[−(x^3 /π)cosπx+((3x^2 )/π)sinπx−((6x)/π^3 )cosπx+(6/π^4 )sinπx]_0 ^1   =1−(6/π^2 )  Var(x)=E(x^2 )−E^2 (x)=1−(6/π^2 )−(1−(4/π^2 ))^2   =(2/π^2 )(1−(8/π^2 ))  ✓
01kxsinπxdx=1kπxcosπx+kπ2sinπx]01=1k=πE(x)=01πx2sinπxdx=π(x2πcosπx+2xπ2sinπx+2π2cosπx]01π(1π4π3)=14π2E(x2)=01x3sinπxdx=π[x3πcosπx+3x2πsinπx6xπ3cosπx+6π4sinπx]01=16π2Var(x)=E(x2)E2(x)=16π2(14π2)2=2π2(18π2)
Commented by Spillover last updated on 16/May/23
your right.
yourright.

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