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).
Answer:
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
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