Question

The pdf of a continuous random variable x is given by f(x)={90x^8(1-x),0<x<1,0 otherwise

Find

a.p(x<or=0.5)

b.p(0.25<x<0.5)

Answer 1

f(x)=90x^8(1-x)=90(x^8-x^9). ∫f(x)dx=90(x^9/9-x^10/10)=90x^9(1/9-x/10) between limits {m,n}.

When m=0 and n=1 the integral=1 which is the sum of all probabilities.

The graph of this PDF is very much weighted to the right, so we can expect low probability values for x≤0.5.

(a) ∫f(x)dx for [0,5]=90(0.5^9)(1/9-1/20)=0.010742 approx.

(b) ∫f(x)dx for [0,0.25]=90(0.25^9)(1/9-1/40)=2.956E-5 approx., so for [0.25,0.5] P is 0.010742-0.000030=0.010713 approx.

Comments

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