Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F (x) = 0 x < 0 x2 25 0 ≤ x < 5 1 5 ≤ x Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.)

(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) ].

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