Question

Suppose that f(x) = 3 (8x - x^2)/256 for 0 < x < 8

Determine the following probabilities. Round all of your answers to 3 decimal places (e.g. 98.765).
a. P(X < 5)
b. P(X< 9)
c. P(5 < X < 7)
d. P(X > 3)
e. Determine x such that P(X < x) = 0.95

Answer 1

Answer and Step-by-step explanation: This function is a probability density function of a random variable X:

`f(x)=(3(8x-x^(2)))/(256)`

and to calculate probabilities, we will have to integrate it:

`P(X<x)=\int\limits^x_0 {(3)/(256)(8x-x^(2)) } \, dx`

Solving:

`P(X<x)=(3)/(256)(4x^(2)-(x^(3))/(3) )`

Then:

a. P(X < 5)

`P(X<5)=\int\limits^5_0 {(3)/(256)(8x-x^(2)) } \, dx`

`P(X<5)=(3)/(256)(100-(125)/(3) )`

`P(X<5)=(3)/(256)((300-125)/(3) )`

`P(X<5)=(175)/(256)`

P(X < 5) = 0.683

b. P(X < 9)

Since density function's upper limit is 8, probability of x < 9 is 100% or 1.

So, P(X < 9) = 1

c. P(5 < X < 7)

`P(5<X<7)=\int\limits^7_5 {(3)/(256)(8x-x^(2)) } \, dx`

`P(5<X<7)=(3)/(256)[4(7)^(2)-(7^(3))/(3)-4(5)^(2)+(5^(3))/(3)]`

`P(5<X<7)=(3)/(256)[96-(218)/(3) ]`

`P(5<X<7)=(70)/(256)`

P(5 < X < 7) = 0.273

d. P(X > 3)

`P(X>3)=\int\limits^8_3 {(3)/(256)(8x-x^(2)) } \, dx`

`P(X>3)=(3)/(256)[256-(512)/(3)-36+9 ]`

`P(X>3)=(3)/(256)(229-(512)/(3) )`

`P(X>3)=(175)/(256)`

P (X > 3) = 0.683

e. P(X < x) = 0.95

`0.95=\int\limits^x_0 {(3)/(256)(8x-x^(2)) } \, dx`

`(3)/(256)[(12x^(2)-x^(3))/(3) ] =0.95`

`(12x^(2)-x^(3))/(256)=0.95`

`12x^(2)-x^(3)=243.2`

`-x^(3)+12x^(2)-243.2=0`

Solving this cubic equation, we have three values for x:

x₁ = -3.909

x₂ = 8.992

x₃ = 6.917

The value of x will the one between 0 and 8, which are the limits of the function. So, value of x which gives a probability of 0.95 is x = 6.917.