given the following function:f(x)={( kx 0<X<10 k(20-x) 10<X<20@0 otherwise)┤Find the value of k such that f(x) is a probability density function.
Since f(x) is a probability density function, then:∫_(-∞)^∞〖f(x)dx=〗 1∴1=∫_0^10〖kx dx〗+k∫_10^20〖(20-x)〗 dx=k[1/2 x^2 ]_0^10+k[20x-1/2 x^2 ]_10^20=50k+k[400-200-20+50]=100k→k=1/100Thus:f(x)={( x/100 0<X<10@((20-x))/100 10<X<20@0 otherwise)┤
Its graph is:
total answers (1)
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Since f(x) is a probability density function, then:
∫_(-∞)^∞〖f(x)dx=〗 1
∴1=∫_0^10〖kx dx〗+k∫_10^20〖(20-x)〗 dx
=k[1/2 x^2 ]_0^10+k[20x-1/2 x^2 ]_10^20
=50k+k[400-200-20+50]=100k→k=1/100
Thus:
f(x)={( x/100 0<X<10@((20-x))/100 10<X<20@0 otherwise)┤
Its graph is:

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