To solve this problem, we are given the probability density function (pdf) for $x$ as:

$p(x) = 0.01(15 - x) \quad \text{for} \quad 0 \leq x \leq 10.$

We are tasked with finding the mean and median of $x$.

1. Mean Calculation

The mean of a continuous random variable $x$ with probability density function $p(x)$ over a given interval is given by:

$\mu = \int_{a}^{b} x \cdot p(x) \, dx$

Substituting the values from the given function:

$\mu = \int_0^{10} x \cdot 0.01(15 - x) \, dx$

This requires solving the integral of $x(15 - x)$ over the interval $[0, 10]$.

2. Median Calculation

The median $M$ of a continuous random variable is the value of $x$ such that the cumulative distribution function (CDF) equals 0.5. Mathematically:

$\int_0^M p(x) \, dx = 0.5$

Substitute $p(x) = 0.01(15 - x)$ and solve for $M$:

$\int_0^M 0.01(15 - x) \, dx = 0.5$

I will now compute both the mean and the median.### Results:

1. Mean: The mean of $x$ is approximately $\mu = 4.167$.
2. Median: The median of $x$ is $M = 3.820$ (the relevant solution, as the second solution exceeds the interval [0, 10]).

Let me know if you would like further details or explanations.

5 Related Questions:

1. How is the cumulative distribution function (CDF) related to the probability density function (PDF)?
2. How can we interpret the median in the context of a continuous random variable?
3. What are other methods of finding the mean of a random variable besides integration?
4. How does the shape of a probability distribution affect the mean and median?
5. Can the mean and median be the same for certain probability distributions?

Tip: When calculating integrals involving polynomials, break the problem into smaller parts to simplify the computations.