Question

3. (15 pts) The amount of gravel (in tons) sold by a particular construction supply company in a given week has a probability density function: f(x) = k(1-12) for 0<r<d, where d is the last digit in your student ID number. (If your ID number ends in a 0, please let d = 10.) (a) Find & (b) On average, how much gravel is this company selling in a given week? (e) Suppose that the amount sold in any particular week is independent of any other. Suppose the ocompany makes a profit in a week if they sell at least tons of gravel. What is the probability that the company was profitable at least 7 out of 12 weeks in a particular quarter?

Please use d = 3.

Answer 1

$\textbf{(a)}$

f(x) = k1-r). 0<x<3

So f. dr =1= k1-r) dr =1=k dr-k rdr =1= 3k-9k = 1 JoJo -6k =1→ k=- Jo

$\textbf{(b)}$

On average, the amount of gravel (in tons) the company is selling in a given week

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\int_0^3 x\,f(x)\,dx=\int_0^3 kx(1-x^2)\,dx= k\int_0^3 x\, dx-k\int_0^3 x^3dx= \left ( -\frac 1 6 \right )\left ( \frac{9}{2} \right )-\left ( -\frac 1 6 \right )\left ( \frac{81}{4}\right )=2.625$

$\textbf{(c)}$

Probability that the company makes a profit in a week

= Probability that the company sells at least 5 tons of gravel a week

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\int_{\frac 3 2}^3 f(x)\,dx=\int_{\frac 3 2}^3 k(1-x^2)\,dx= k\int_{\frac 3 2}^3 dx-k\int_{\frac 3 2}^3 x^2dx= \left ( -\frac 1 6 \right )\left ( \frac{3}{2} \right )-\left ( -\frac 1 6 \right )\left ( \frac{63}{8}\right )=1.0625$

This is not possible. There must be something wrong in the question.