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

Solution :

Given :

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

1. Cumulative distribution function

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

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

             $=0+\int_6^x \frac{1}{450}(x^2-x) \ dx$

             $=\frac{1}{450} \int_6^x (x^2-x) \ dx$

             $=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$

             $=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right]$

            $=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$

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

                       $=\int_{6}^{12}x . \left( \frac{1}{450} \ (x^2-x)\right)\ dx$

                     $=\frac{1}{450} \int_6^{12} \ (x^3 - x^2) \ dx$

                     $=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$

                     $=\frac{1}{450} \left[ \left(\frac{(12)^4}{4} - \frac{(12)^3}{3} \right) - \left(\frac{(6)^4}{4} - \frac{(6)^3}{3} \right)$

                     $=\frac{1}{450} [4608 - 252]$

                    = 17.2857