quality control If the probability density of a random varia

If the probability density of a random variable is given by f(x) = {k(1 -x^2) for 0

Solution

a)

Normalizing,

Integral [k(1-x^2) dx]|(0,1) = 1

k(x - x^3/3)|(0,1) = 1

k(2/3) = 1

k = 3/2 or 1.5 [ANSWER]

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b)

Thus, the cumulative distribution function here is

P(X<x) = F(x) = 1.5(x - x^3/3) , 0<x<1
0 , elsewhere

Thus,

P(0.1<x<0.2) = F(0.2) - F(0.1) = 1.5*(0.2 - 0.2^3/3) - 1.5*(0.1 - 0.1^3/3)

P(0.1<x<0.2) = 0.1465 [ANSWER]

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c)

P(x>0.5) = 1 - F(0.5) = 1 - 1.5*(0.5 - 0.5^3/3)

= 0.3125 [ANSWER]