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The problem describes a discrete random variable \( M \) with the probability distribution:

\[
f(m) = \begin{cases}
\frac{m}{36}, & m = 1, 2, 3, \dots, 8 \\
0, & \text{elsewhere}
\end{cases}
\]

### Step 1: Finding the Mean (Expected Value) \( \mu \)

The expected value \( \mu \) of a discrete random variable is given by:

\[
\mu = E[M] = \sum_{m} m \cdot f(m)
\]

Using the given distribution, \( f(m) = \frac{m}{36} \) for \( m = 1, 2, \dots, 8 \), the mean is:

\[
\mu = \sum_{m=1}^{8} m \cdot \frac{m}{36} = \frac{1}{36} \sum_{m=1}^{8} m^2
\]

First, compute the sum of squares:

\[
\sum_{m=1}^{8} m^2 = 1^2 + 2^2 + 3^2 + \dots + 8^2 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204
\]

So the expected value is:

\[
\mu = \frac{1}{36} \times 204 = \frac{204}{36} = 5.67
\]

### Step 2: Finding the Variance \( \sigma^2 \)

The variance is given by:

\[
\sigma^2 = E[M^2] - (E[M])^2
\]

First, we compute \( E[M^2] \):

\[
E[M^2] = \sum_{m=1}^{8} m^2 \cdot f(m) = \sum_{m=1}^{8} m^2 \cdot \frac{m}{36} = \frac{1}{36} \sum_{m=1}^{8} m^3
\]

Now, compute the sum of cubes:

\[
\sum_{m=1}^{8} m^3 = 1^3 + 2^3 + 3^3 + \dots + 8^3 = 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 = 1296
\]

Thus, \( E[M^2] \) is:

\[
E[M^2] = \frac{1}{36} \times 1296 = 36
\]

Finally, the variance is:

\[
\sigma^2 = 36 - (5.67)^2 = 36 - 32.1489 = 3.8511
\]

### Summary:
- **Mean \( \mu \) = 5.67**
- **Variance \( \sigma^2 \) = 3.8511**

Would you like further clarification or details on any step?

Here are 5 related questions to expand on this:
1. How would the mean and variance change if the probability distribution were uniform?
2. What is the standard deviation of \( M \) based on the variance we found?
3. How would the calculations differ if the range of \( m \) values increased to 10?
4. Can you verify the calculation of \( E[M^2] \) using a different method?
5. How would you interpret these results in terms of real-life applications?

**Tip:** When calculating variance, always check both \( E[M^2] \) and \( E[M]^2 \) separately to ensure accuracy.

A discrete random variable M has the probability distribution f(m) = m/36 for m = 1, 2, 3, ..., 8. Find the mean and variance of M.

Solve a probability problem involving a discrete random variable M with the distribution f(m) = m/36 for m = 1, 2, ..., 8. Learn how to calculate the mean (expected value) and variance using key probability formulas and techniques. Perfect for undergraduate students studying probability.

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