11. Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x) 5 5 0 x , 0 x2 4 0 # x , 2 1 2 # x a. Calculate P(X # 1). b. Calculate P(.5 # X # 1). c. Calculate P(X . 1.5). d. What is the median checkout duration m ,? [solve .5 5 F(m ,)]. e. Obtain the density function f (x). f. Calculate E(X).

Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 x<0

x²/25 0 ≤ x ≤ 5

1 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

109 is the nearest whole number/final answer! : )