Solved: The probability that ore samples taken from a region contain between and iron is

Chapter 5, Problem 106

(choose chapter or problem)

In Exercises 105 and 106, the function

\(f(x)=k x^{n}(1-x)^{m}, \quad 0 \leq x \leq 1\)

where \(n>0, m>0\), and k is a constant, can be used to represent various probability distributions. If k is chosen such that

\(\int_{0}^{1} f(x) d x=1\)

then the probability that x will fall between a and b \((0 \leq a \leq b \leq 1)\) is

\(P_{a, b}=\int_{a}^{b} f(x) d x\)

The probability that ore samples taken from a region contain between 100a% and 100b% iron is

\(P_{a, b}=\int_{a}^{b} \frac{1155}{32} x^{3}(1-x)^{3 / 2} d x\)

where x represents the proportion of iron. (See figure.) What is the probability that a sample will contain between

(a) 0% and 25% iron?

(b) 50% and 100% iron?

Text Transcription:

f(x)=k x^n(1-x)^m, 0 leq x leq 1

n>0, m>0

int_0^1 f(x) d x=1

(0 leq a leq b leq 1)

P_a, b=int_a^b f(x) d x

P_a, b=int_a^b frac 1155 32 x^3(1-x)^3 / 2 d x