The probability density function of x is given as f(x) = ae^-x/5 for x > 0. The value of a is ? (a) 0·5 (b) 0·3 (c) 0·2 (d) 0·1

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Thank you. Please help. Is it by taking the integral value = 1?

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Answer 1 · 2017-12-03T16:39:09

#a=0.2#

pdf#" "f(x)=ae^(-x/5), a>0#

for a pdf

#int_(all" " x)f(x)=1#

#:.int_0^ooae^(-x/5)dx=1#

#Lim_(trarroo)int_0^tae^(-x/5)dx=1#

#Lim_(trarroo)[-5ae^(-x/5)]_0^t=1#

#Lim_(trarroo)[-5ae^(-x/5)]^t-[-5ae^(-x/5)]_0=1#

#0--5ae^0=1#

#=>5a=1#

#:.a=1/5=0.2#