We are given a probability density function (PDF) $p(x) = Cx(4 - x)$ on the interval $[0, 4]$, and the task is to find the constant $C$ such that $p(x)$ is a valid PDF, and then compute the probability $P(3 \leq X \leq 4)$.

Step 1: Find the constant $C$

For $p(x)$ to be a valid probability density function, the integral of $p(x)$ over the given interval must equal 1. This is the normalization condition:

$\int_0^4 p(x) \, dx = 1$

Substituting $p(x) = Cx(4 - x)$:

$\int_0^4 Cx(4 - x) \, dx = 1$

First, expand the expression inside the integral:

$Cx(4 - x) = C(4x - x^2)$

Now, integrate term by term:

$\int_0^4 C(4x - x^2) \, dx = C \left[ \int_0^4 4x \, dx - \int_0^4 x^2 \, dx \right]$

Compute the individual integrals:

$\int_0^4 4x \, dx = \left[ 2x^2 \right]_0^4 = 2(4^2) - 2(0^2) = 32$

$\int_0^4 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^4 = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3}$

Now substitute these values back into the equation:

$C \left( 32 - \frac{64}{3} \right) = 1$

Simplify the expression inside the parentheses:

$32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$

Thus, we have:

$C \times \frac{32}{3} = 1$

Solve for $C$:

$C = \frac{3}{32}$

Step 2: Compute $P(3 \leq X \leq 4)$

Now that we have found $C = \frac{3}{32}$, we can compute the probability $P(3 \leq X \leq 4)$:

$P(3 \leq X \leq 4) = \int_3^4 p(x) \, dx = \int_3^4 \frac{3}{32} x(4 - x) \, dx$

First, expand $x(4 - x)$:

$x(4 - x) = 4x - x^2$

Thus, the integral becomes:

$P(3 \leq X \leq 4) = \frac{3}{32} \int_3^4 (4x - x^2) \, dx$

Now, integrate term by term:

$\int_3^4 4x \, dx = \left[ 2x^2 \right]_3^4 = 2(4^2) - 2(3^2) = 2(16) - 2(9) = 32 - 18 = 14$

$\int_3^4 x^2 \, dx = \left[ \frac{x^3}{3} \right]_3^4 = \frac{4^3}{3} - \frac{3^3}{3} = \frac{64}{3} - \frac{27}{3} = \frac{37}{3}$

Substitute these values back into the integral:

$P(3 \leq X \leq 4) = \frac{3}{32} \left( 14 - \frac{37}{3} \right)$

Simplify the expression inside the parentheses:

$14 - \frac{37}{3} = \frac{42}{3} - \frac{37}{3} = \frac{5}{3}$

Thus, the probability becomes:

$P(3 \leq X \leq 4) = \frac{3}{32} \times \frac{5}{3} = \frac{5}{32}$

Final Answer:

The constant $C$ is $\frac{3}{32}$, and the probability $P(3 \leq X \leq 4) = \frac{5}{32}$.

Would you like further details or have any questions about this process?

Related Questions:

1. How do you verify if a given function is a valid probability density function?
2. What is the significance of normalization in probability density functions?
3. How do you calculate the expected value of a probability density function?
4. What happens if the integral of a PDF does not equal 1?
5. How would you compute a probability for a continuous distribution with a different interval?

Tip: Always ensure that the total probability for a PDF over its entire range sums to 1. If it doesn't, the function isn't valid as a PDF.