Math Problem Statement

Find a constant C such that p(x) is a probability density function on the given interval, and compute the probability indicated: p(x) = C / (x+4)^3 on [0, ∞); P(0 ≤ X ≤ 1).

Solution

To solve the problem:

  1. Find the constant CC so that p(x)p(x) is a valid probability density function. The condition for a probability density function is that the total integral over the given range must equal 1. The function given is:

    p(x)=C(x+4)3p(x) = \frac{C}{(x + 4)^3}

    on the interval [0,)[0, \infty). We need to compute:

    0C(x+4)3dx=1\int_0^\infty \frac{C}{(x+4)^3} dx = 1

    This will allow us to solve for CC.

  2. Compute P(0X1)P(0 \leq X \leq 1) by integrating the probability density function over the interval from 0 to 1:

    P(0X1)=01C(x+4)3dxP(0 \leq X \leq 1) = \int_0^1 \frac{C}{(x+4)^3} dx


Step 1: Find CC

First, compute the integral:

0C(x+4)3dx=1\int_0^\infty \frac{C}{(x+4)^3} dx = 1

This is a standard integral. Using the substitution u=x+4u = x + 4, we can simplify the integration:

01(x+4)3dx=41u3du\int_0^\infty \frac{1}{(x+4)^3} dx = \int_4^\infty \frac{1}{u^3} du

The integral of u3u^{-3} is:

1u3du=12u2\int \frac{1}{u^3} du = \frac{1}{2u^2}

Evaluating the limits at u=4u = 4 and uu \to \infty:

[12u2]4=01216=132\left[ \frac{-1}{2u^2} \right]_4^\infty = 0 - \frac{-1}{2 \cdot 16} = \frac{1}{32}

Thus:

C132=1C=32C \cdot \frac{1}{32} = 1 \quad \Rightarrow \quad C = 32

Step 2: Compute P(0X1)P(0 \leq X \leq 1)

Now we compute the probability for 0X10 \leq X \leq 1:

P(0X1)=0132(x+4)3dxP(0 \leq X \leq 1) = \int_0^1 \frac{32}{(x+4)^3} dx

Using the same substitution u=x+4u = x + 4, the limits change from u=4u = 4 to u=5u = 5, and we have:

0132(x+4)3dx=32451u3du\int_0^1 \frac{32}{(x+4)^3} dx = 32 \int_4^5 \frac{1}{u^3} du

Again, the integral of u3u^{-3} is 12u2\frac{-1}{2u^2}, so:

32[12u2]45=32(12251216)32 \left[ \frac{-1}{2u^2} \right]_4^5 = 32 \left( \frac{-1}{2 \cdot 25} - \frac{-1}{2 \cdot 16} \right)

Simplifying:

32(150+132)=32(16+25800)=329800=288800=0.3632 \left( \frac{-1}{50} + \frac{1}{32} \right) = 32 \cdot \left( \frac{-16 + 25}{800} \right) = 32 \cdot \frac{9}{800} = \frac{288}{800} = 0.36


Final Answers:

  • C=32C = 32
  • P(0X1)=0.36P(0 \leq X \leq 1) = 0.36

Would you like further details or have any questions? Here are some follow-up questions to deepen your understanding:

  1. What is the significance of the constant CC in a probability density function?
  2. How does substitution simplify the integration of functions involving transformations like x+4x+4?
  3. How do the properties of definite integrals ensure the probability is valid over infinite intervals?
  4. Could you extend the calculation for different intervals, say P(1X5)P(1 \leq X \leq 5)?
  5. Why is it important that the total probability over the interval [0,)[0, \infty) equals 1?

Tip: When dealing with probability density functions, always ensure the integral over the entire range equals 1 to validate the function.

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Math Problem Analysis

Mathematical Concepts

Probability Density Function
Integration
Substitution Method

Formulas

p(x) = C / (x + 4)^3
∫_0^∞ (C / (x+4)^3) dx = 1
P(0 ≤ X ≤ 1) = ∫_0^1 (C / (x+4)^3) dx

Theorems

Definite Integrals
Properties of Probability Density Functions

Suitable Grade Level

Undergraduate