Question

Suppose the c.d.f. of X is F(t) 3 for 0<t< (a) What is F(5)? (b) What is F(-5)? (c) Compute the p.d.f of X. (d) Compute the mean of X (e) Compute the variance of X. (f) Compute the standard deviation of X (g) Compute the squared coefficient of variation of X.

Answer 1

a)F(5)=1 as 5 is above upper bound 1.

b) F(-5)=0 as -5 is below lower bound of 0

c)

pdf fo X : f(x) =(d/dx)F(x)=(d/dx)*t³ =3t²

d)

mean of X =E(X)= $\int_{0}^{1}$ t*f(t) dt = $\int_{0}^{1}$ 3t³ dt =3t⁴/4 |¹₀ =3/4

e)

E(X²)= $\int_{0}^{1}$ t²*f(t) dt = $\int_{0}^{1}$ 3t⁴ dt =3t⁵/5 |¹₀ =3/5

Variance of X =Var(X)=E(X²)-(E(X))² =3/5-(3/4)² =3/80

f)

std deviaiton of X =sqrt(Var(x))=sqrt(3/80)

g)squared coefficent of variation =Var(x)/(E(X))² =(3/80)/(9/16)=1/15 or 6.67%