To find the mean value $E(X)$ of the random variable $X$ given its probability density function (pdf), we use the formula:

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

Here, the pdf $f(x)$ is given as:

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

Since the function is defined only in the interval $0 \leq x \leq 100$, the mean becomes:

$E(X) = \int_0^{100} x \cdot \frac{3}{5} \times 10^{-5} \cdot x (100 - x) \, dx$

Step-by-step solution:

1. Simplify the integrand:

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

2. Expand $x^2 (100 - x)$:

$x^2 (100 - x) = 100x^2 - x^3$

So, the integral becomes:

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

3. Break the integral into two parts:

$E(X) = \frac{3}{5} \times 10^{-5} \left( \int_0^{100} 100x^2 \, dx - \int_0^{100} x^3 \, dx \right)$

4. Compute the integrals:

$\int_0^{100} 100x^2 \, dx = 100 \times \frac{x^3}{3} \Bigg|_0^{100} = 100 \times \frac{100^3}{3} = \frac{100^4}{3} = \frac{10^8}{3}$

$\int_0^{100} x^3 \, dx = \frac{x^4}{4} \Bigg|_0^{100} = \frac{100^4}{4} = \frac{10^8}{4}$

5. Substitute the results:

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

6. Find a common denominator:

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

7. Simplify:

$E(X) = \frac{3}{5} \times 10^{-5} \times \frac{10^8}{12}$

$E(X) = \frac{3}{5} \times \frac{10^3}{12} = \frac{3 \times 10^3}{5 \times 12} = \frac{3000}{60} = 50$

Final Answer:

The mean value of the random variable is $\boxed{50}$.

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Would you like to explore more details about the calculation or have any specific questions? Here are 5 related questions for deeper understanding:

1. How would you calculate the variance of this random variable?
2. What is the significance of the mean in the context of this distribution?
3. How does changing the range of $x$ affect the mean?
4. Can you derive the general formula for the mean for a quadratic pdf like this one?
5. What is the probability that $X$ lies between two values, say 25 and 75?

Tip: The mean value provides the central tendency of the distribution, offering a first insight into where most values of the random variable are likely to occur.