Let X be a continuous random variable with density function

Let X be a continuous random variable with density function f_x (x) = 1/2 x^-3/2 1

Solution

a) P(X > 10) = 1 - P (X < 10)

P(X < 10) = integral of f(x) = 1/2 x^-3/2

Integrating we get [1/2 x^-1/2 / (-1/2) ]

= [x^-1/2] from 10 to 1

= [1 - 1/sqrt(10)]

P (X<10) =  [1 - 1/sqrt(10)]

Hence, required probability = 1 -[1 - 1/sqrt(10)]

P (X > 10) = 1/ sqrt(10)

= 0.3162

b)

Mean = integral of xf(x)

Thus, integrating xf(x) = 1/2 x^-1/2

= [1/2 x^1/2 / (1/2)]

= sqrt(x) from 1 to infinity

= infinity

Thus, the mean of the pdf function is unbounded.

Hope this helps.