Question

The thickness X of aluminum sheets is distributed according to the probability density function f(x) = 450 (x2 - x) if 6 < x < 12 0 otherwise 5-1 Derive the cumulative distribution function F(x) for 6 < x < 12. The answer is a function of x and is NOT 1! Show the antiderivative in your solution. 5-2 What is E(X) = {the mean of all sheet thicknesses)? Show the antiderivative in your solution.

Answer 1

Given :

`f(x) = \left\{\begin{matrix}(1)/(450)(x^2-x) & \text{if } 6 < x < 12 \\ 0 & \text{otherwise}\end{matrix}\right.`

1. Cumulative distribution function

`$P(X \leq x) = \int_(- \infty)^x f(x) \ dx$`

`$=\int_(- \infty)^6 f(x) dx + \int_(6)^x f(x) dx $`

`$=0+\int_6^x (1)/(450)(x^2-x) \ dx$`

`$=(1)/(450) \int_6^x (x^2-x) \ dx$`

`$=(1)/(450)\left[(x^3)/(3)-(x^2)/(2)\right]_6^x$`

`$=(1)/(450)\left[ \left( (x^3)/(3) - (x^2)/(2)\left) - \left( (6^3)/(3) - (6^2)/(2) \right) \right] $`

`$=(1)/(450)\left[(x^3)/(3) - (x^2)/(2) - 54 \right]$`

2. Mean
`$E(x) = \int_(- \infty)^(\infty) \ x \ f(x) \ dx$`

`$=\int_(6)^(12)x . \left( (1)/(450) \ (x^2-x)\right)\ dx$`

`$=(1)/(450) \int_6^(12) \ (x^3 - x^2) \ dx$`

`$=(1)/(450) \left[(x^4)/(4) - (x^3)/(3) \right]_6^(12) \ dx$`

`$=(1)/(450) \left[ \left(((12)^4)/(4) - ((12)^3)/(3) \right) - \left(((6)^4)/(4) - ((6)^3)/(3) \right) $`

`$=(1)/(450) [4608 - 252]$`

= 17.2857