Et 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 16 0 ≤ x < 4 1 4 ≤ x Use the cdf to obtain the following:

(a) P (X %u2264 1)

(b) P (0.5 %u2264 X %u2264 1)

(c) P (X > 1.5)

(d) The median checkout duration [solve 0.5 = F (mew) ]

(e) Use F' (x) to obtain the density function f (x)

(f) Calculate E (X)

(g) Calculate V (X) and %u03C3x

(h) If the borrower is charged an amount h (X) = X2 when checkout duration is X, compute the expected charge

a) P (x < = 1) = 0.0625

b) P (0.5 < = x < = 1) = 0.04688

c) P (x > 1.5) = 0.8594

d) x = 2.8284

e) f (x) = x / 8

f) E (X) = 2.6667

g) Var (X) = 0.8888, s. d (X) = 0.9428

h) E[h (X) ] = 256

Step-by-step explanation:

Given:

The cdf is as follows:

F (x) = 0 x < 0

F (x) = (x^2 / 16) 0 < x < 4

F (x) = 1 x > 4

Find:

(a) Calculate P (X ≤ 1).

(b) Calculate P (0.5 ≤ X ≤ 1).

(c) Calculate P (X > 1.5).

(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 < 1.

So, P (x < = 1) = (x^2 / 16) | 0 to 1

P (x < = 1) = (1^2 / 16) - 0

P (x < = 1) = 0.0625

b) Evaluate the cdf given with the limits 0.5 < x < 1.

So, P (0.5 < = x < = 1) = (x^2 / 16) | 0.5 to 1

P (0.5 < = x < = 1) = (1^2 / 16) - (0.5^2 / 16)

P (0.5 < = x < = 1) = 0.0625 - 0.015625 = 0.04688

c) Evaluate the cdf given with the limits x > 1.5

So, P (x > 1.5) = 1 - P (x < = 1.5)

P (x > 1.5) = 1 - (1.5^2 / 16) - 0

P (x > 1.5) = 1 - 0.140625 = 0.8594

d) The median checkout for the duration that is 50% of the probability:

So, P (x < a) = 0.5

(x^2 / 16) = 0.5

x^2 = 12.5

x = 2.8284

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 / 8

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 / 8)

E (X) = x^3 / 24 limits: 0 to 4

E (X) = 4^3 / 24 = 2.6667

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 / 8). dx - (E (X)) ^2

Var (X) = x^4 / 32 | - (2.666667) ^2 limits: 0 to 4

Var (X) = 4^4 / 32 - (2.666667) ^2 = 0.88888

s. d (X) = sqrt (Var (X)) = sqrt (0.88888) = 0.9428

h) Find the expected charge E[h (X) ], where h (X) is given by:

h (x) = (f (x)) ^2 = x^2 / 64

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 / 64)

E (h (X))) = x^4 / 256 limits: 0 to 16

E (h (X))) = 16^4 / 256 = 256