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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

Respuesta :

Answer:

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

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