The probability density function (p.d.f.) of a continuous random variable X is defined to be:
f(x)= x/6+k for 0<x<2.50 otherwise,
for some constant k.
For these problems, please ensure your answers are accurate to within 3 decimals.
Part a) Find the value of k that makes the above function a proper p.d.f.
Part b) Hence find P(0.5<X<1).
Answer:
a. k = 0.192
b. P(0.5<X<1) = 1
Step-by-step explanation:
Given
f(x) = x/6 + k for 0 x < 2.50
To find the value of k that makes the above function a proper p.d.f.
∫ f(x) dx must be equal to 1
∫ f(x) dx = 1
Substitute f(x) = x/6 + k
∫ x/6 + k dx {0,2.50}
Integrate with respect to x
x²/(6*2) + kx {0,2.50} = 1
x²/12 + kx {0,2.50} = 1
(2.50²/12 + 2.50k) - (0²/12 - 0*k) = 1
6.25/12 + 2.50k = 1
Collect like terms
2.50k = 1 - 6.25/12
2.50k = (12 - 6.25)/12
2.50k = 5.75/12
Divide through by 2.50
2.50k/2.50 = 5.75/12 * 1/2.50
k = 2.3/12
k = 23/120
k = 0.192 --- Approx to 3 decimal places
So f(x) = x/6 + 23/120 for 0 x < 2.50
b.
Find P(0.5<X<1)
Given that f(x) = x/6 + 23/120 for 0 x < 2.50
P(0.5<X<1) = ∫x/6 + 23/120 {0,2.50}
P(0.5<X<1) = x²/12+ 23x/120 {0,2.50}
P(0.5<X<1) = (2.5²/12+ 23*2.5/120) - (0²/12+ 23*0/120)
P(0.5<X<1) = (2.5²/12+ 23*2.5/120) - (0)
P(0.5<X<1) = (6.25/12+ 57.5/120) - (0)
P(0.5<X<1) = (6.25/12+ 5.75/12)
P(0.5<X<1) = (6.25 + 5.75)/12
P(0.5<X<1) = 12/12
P(0.5<X<1) = 1
