Question

A deposit has the grade attribute $X \in [0,30]$ with a density function $f(x)$. For a cut-off grade $x_c$, the proportion of the ore in the deposit is given by

• A. $\displaystyle \int_{0}^{30} f(x)dx - \int_{0}^{x_c} f(x)dx$
• B. $\displaystyle \frac{1}{2}\int_{0}^{30} f(x)dx - \int_{0}^{x_c} f(x)dx$
• C. $\displaystyle \frac{1}{2}\int_{0}^{30} f(x)dx + \int_{0}^{x_c} f(x)dx$
• D. $\displaystyle \int_{0}^{x_c} f(x)dx$

Hint:

Ore proportion above cutoff is simply $P(X>x_c)$, computed as the tail area of the PDF.

Answer 1

The Correct Option is D

The variable $X$ represents grade with support on $[0,30]$.
A cut-off grade $x_c$ means ore with grade greater than $x_c$ is considered economic ore.
Thus, proportion of ore = $P(X>x_c)$.
\[ P(X>x_c) = \int_{x_c}^{30} f(x)\,dx. \]
Using total probability:
\[ \int_{x_c}^{30} f(x)dx = \int_{0}^{30} f(x)dx - \int_{0}^{x_c} f(x)dx. \]
If $f(x)$ is a valid density on $[0,30]$, then $\int_0^{30} f(x)dx = 1$, but the expression in option (A) is general and correct.
Final Answer: (A)