Question

The continuous random variable X represents the percentage of iron contained in an ore sample taken from a particular region. The probability density function for X is given by f(x) = kx^3 (1 – x) where k is a positive constant.
Required:
a. On what interval does X take values?
b. Find k.
c. What is the probability that an ore sample contains at least 70% iron?

Answer 1

Answer:

a) X take values in the interval [0; 1].

b) k = 20

c) P(X≥0.7)=0.46

Step-by-step explanation:

As X represents percentage of iron contained in an ore sample taken from a particular region, it can take values within 0 (0%) and 1 (100%).

Within this values (0 and 1), to be a valid probability density function, the area under the curve of the density probability function has to be 1. This will allow us to calculate the value of k.

This can be written as:

$\int\limits^1_0 {kx^3(1-x)} \, dx=1\\\\\\k \int\limits^1_0 {(x^3-x^4)} \, dx=1\\\\\\k\left(\dfrac{x^4}{4}-\dfrac{x^5}{5} \right) |_{0;1}=1\\\\\\k\left(\dfrac{1^4}{4}-\dfrac{1^5}{5} \right)-k \left(\dfrac{0^4}{4}-\dfrac{0^5}{5} \right)=1\\\\k\cdot \dfrac{5-4}{20}=1\\\\\\k\cdot \dfrac{1}{20}=1\\\\\\k=20$

Now, we have to calculate the probability that x≥0.7 (ore sample contains at least 70% iron).

This can be calculated as:

$P(x\geq0.7)=\int\limits^1_{0.7} {20x^3 (1-x)} \, dx \\\\\\P(x\geq0.7)=20\left(\dfrac{1^4}{4}-\dfrac{1^5}{5}\right)-20\left(\dfrac{0.7^4}{4}-\dfrac{0.7^5}{5}\right)\\\\\\P(x\geq0.7)=1-20(0.060-0.033)=1-0.54=0.46$