Answer:
a) P (x <= 3 ) = 0.36
b) P ( 2.5 <= x <= 3 ) = 0.11
c) P (x > 3.5 ) = 1 - 0.49 = 0.51
d) x = 3.5355
e) f(x) = x / 12.5
f) E(X) = 3.3333
g) Var (X) = 13.8891 , s.d (X) = 3.7268
h) E[h(X)] = 2500
Step-by-step explanation:
Given:
The cdf is as follows:
F(x) = 0 x < 0
F(x) = (x^2 / 25) 0 < x < 5
F(x) = 1 x > 5
Find:
(a) Calculate P(X ≤ 3).
(b) Calculate P(2.5 ≤ X ≤ 3).
(c) Calculate P(X > 3.5).
(d) What is the median checkout duration ? [solve 0.5 = F()].
(e) Obtain the density function f(x). f(x) = F '(x) =
(f) Calculate E(X).
(g) Calculate V(X) and σx. V(X) = σx =
(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].
Solution:
a) Evaluate the cdf given with the limits 0 < x < 3.
So, P (x <= 3 ) = (x^2 / 25) | 0 to 3
P (x <= 3 ) = (3^2 / 25) - 0
P (x <= 3 ) = 0.36
b) Evaluate the cdf given with the limits 2.5 < x < 3.
So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3
P ( 2.5 <= x <= 3 ) = (3^2 / 25) - (2.5^2 / 25)
P ( 2.5 <= x <= 3 ) = 0.36 - 0.25 = 0.11
c) Evaluate the cdf given with the limits x > 3.5
So, P (x > 3.5 ) = 1 - P (x <= 3.5 )
P (x > 3.5 ) = 1 - (3.5^2 / 25) - 0
P (x > 3.5 ) = 1 - 0.49 = 0.51
d) The median checkout for the duration that is 50% of the probability:
So, P( x < a ) = 0.5
(x^2 / 25) = 0.5
x^2 = 12.5
x = 3.5355
e) The probability density function can be evaluated by taking the derivative of the cdf as follows:
pdf f(x) = d(F(x)) / dx = x / 12.5
f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:
E(X) = integral ( x . f(x)).dx limits: - ∞ to +∞
E(X) = integral ( x^2 / 12.5)
E(X) = x^3 / 37.5 limits: 0 to 5
E(X) = 5^3 / 37.5 = 3.3333
g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:
Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2 limits: - ∞ to +∞
Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2
Var(X) = x^4 / 50 | - (3.3333)^2 limits: 0 to 5
Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891
s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268
h) Find the expected charge E[h(X)] , where h(X) is given by:
h(x) = (f(x))^2 = x^2 / 156.25
The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:
E(h(X))) = integral ( x . h(x) ).dx limits: - ∞ to +∞
E(h(X))) = integral ( x^3 / 156.25)
E(h(X))) = x^4 / 156.25 limits: 0 to 25
E(h(X))) = 25^4 / 156.25 = 2500
is the probability of both A and B happening.
out of 0.5(fair coin).