To find the mean value $\mu$ of a random variable with a given probability density function (PDF), we use the formula:

$\mu = E[X] = \int_{-\infty}^{\infty} x f(x) dx$

For the given problem, the PDF $f(x)$ is provided as:

$f(x) = \frac{3}{5} \times 10^5 \times x(100 - x), \quad 0 \leq x \leq 100$

Since the PDF is zero outside the range $[0, 100]$, we only need to integrate within that range. So, the formula becomes:

$\mu = \int_0^{100} x f(x) dx$

Substitute the given PDF $f(x)$ into the integral:

$\mu = \int_0^{100} x \left( \frac{3}{5} \times 10^5 \times x(100 - x) \right) dx$

Simplify the expression inside the integral:

$\mu = \frac{3 \times 10^5}{5} \int_0^{100} x^2 (100 - x) dx$

Now, distribute $x^2$ through the term $(100 - x)$:

$\mu = \frac{3 \times 10^5}{5} \int_0^{100} (100x^2 - x^3) dx$

Next, compute the integral term by term:

1. The integral of $100x^2$: $\int_0^{100} 100x^2 dx = 100 \left[ \frac{x^3}{3} \right]_0^{100} = 100 \times \frac{100^3}{3} = \frac{100^4}{3} = \frac{10^8}{3}$

2. The integral of $x^3$: $\int_0^{100} x^3 dx = \left[ \frac{x^4}{4} \right]_0^{100} = \frac{100^4}{4} = \frac{10^8}{4}$

Now substitute these results back into the expression for $\mu$:

$\mu = \frac{3 \times 10^5}{5} \left( \frac{10^8}{3} - \frac{10^8}{4} \right)$

Simplify the terms inside the parentheses:

$\frac{10^8}{3} - \frac{10^8}{4} = \frac{4 \times 10^8 - 3 \times 10^8}{12} = \frac{10^8}{12}$

Thus:

$\mu = \frac{3 \times 10^5}{5} \times \frac{10^8}{12} = \frac{3 \times 10^5 \times 10^8}{5 \times 12} = \frac{3 \times 10^{13}}{60} = 50$

So, the mean value of the random variable is indeed $\mu = 50$.

Would you like further details on any step?

Here are 5 related questions you could explore:

1. How do you find the variance of a random variable given its PDF?
2. What is the significance of the mean value in probability theory?
3. How do different types of distributions (normal, uniform, etc.) affect the mean value?
4. What role does normalization play in ensuring a valid PDF?
5. How would you approach finding the median value for this PDF?

Tip: Always check that the total probability (integral of the PDF over the range) equals 1 to confirm it's a valid PDF.