The mean marks of learners in Sugakope PCG Basic is 2.5. Use the table below to answer the following questions: a. Find the mean mark, b. modal mark, c. median.

The problem states that the mean mark is 2.5. Let's use this information to find the value of $m$.

The formula for the mean of grouped data is:

$$\bar{x} = \frac{\sum (x \cdot f)}{\sum f}$$

First, calculate the sum of frequencies ($\sum f$) and the sum of (mark $\times$ frequency) ($\sum (x \cdot f)$) in terms of $m$.

Marks ($x$): 1, 2, 3, 4, 5
Frequency ($f$): 3, $m$, 5, $m+2$, 3

Step 1: Calculate the total frequency and the sum of (mark $\times$ frequency).

$$\sum f = 3 + m + 5 + (m+2) + 3 = 13 + 2m$$ $$\sum (x \cdot f) = (1 \cdot 3) + (2 \cdot m) + (3 \cdot 5) + (4 \cdot (m+2)) + (5 \cdot 3)$$ $$\sum (x \cdot f) = 3 + 2m + 15 + 4m + 8 + 15$$ $$\sum (x \cdot f) = 41 + 6m$$

Step 2: Use the given mean to solve for $m$.
Given $\bar{x} = 2.5$:

$$2.5 = \frac{41 + 6m}{13 + 2m}$$ $$2.5(13 + 2m) = 41 + 6m$$ $$32.5 + 5m = 41 + 6m$$ $$32.5 - 41 = 6m - 5m$$ $$m = -8.5$$

Since frequency cannot be negative, there is an inconsistency in the problem statement. It is not possible to find a valid positive integer value for $m$ that results in a mean of 2.5 with the given frequency distribution.

To provide a complete solution, I will proceed by assuming a plausible positive integer value for $m$. Let's assume $m=1$. This is a common approach when a problem is ill-posed in a test setting.

Step 3: Recalculate frequencies and total frequency with the assumption $m=1$.
Marks ($x$): 1, 2, 3, 4, 5
Frequency ($f$): 3, 1, 5, (1+2)=3, 3
Total frequency $\sum f = 3 + 1 + 5 + 3 + 3 = 15$.

Step 4: Recalculate $\sum (x \cdot f)$ with $m=1$.

$$\sum (x \cdot f) = (1 \cdot 3) + (2 \cdot 1) + (3 \cdot 5) + (4 \cdot 3) + (5 \cdot 3)$$ $$\sum (x \cdot f) = 3 + 2 + 15 + 12 + 15 = 47$$