For the following distribution of marks obtained by 50 students, find the mid-values of each class and state the coordinates needed to plot a frequency polygon.

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Number of students | 5 | 8 | 12 | 15 | 10

Question

For the following distribution of marks obtained by 50 students, find the mid-values of each class and state the coordinates needed to plot a frequency polygon.

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Number of students | 5 | 8 | 12 | 15 | 10 |

Accepted Answer

**Step 1 — compute mid-values** using $	ext{mid-value} = \dfrac{	ext{lower limit} + 	ext{upper limit}}{2}$:

| Class | Mid-value | Frequency |
|---|---|---|
| 0–10 | $\frac{0+10}{2}=5$ | 5 |
| 10–20 | $\frac{10+20}{2}=15$ | 8 |
| 20–30 | $\frac{20+30}{2}=25$ | 12 |
| 30–40 | $\frac{30+40}{2}=35$ | 15 |
| 40–50 | $\frac{40+50}{2}=45$ | 10 |

**Step 2 — plot the points.** (5, 5), (15, 8), (25, 12), (35, 15), (45, 10).

**Step 3 — close the polygon.** Since there is no class before 0–10 or after 40–50, an imaginary class of the same width (width 10) is added on each side with frequency 0: a class $-10$ to $0$ with mid-value $-5$, and a class $50$ to $60$ with mid-value $55$. The polygon is therefore joined down to the x-axis at $(-5, 0)$ and $(55, 0)$. …

Class	Mid-value	Frequency
0–10	$\frac{0+10}{2}=5$	5
10–20	$\frac{10+20}{2}=15$	8
20–30	$\frac{20+30}{2}=25$	12
30–40	$\frac{30+40}{2}=35$	15
40–50	$\frac{40+50}{2}=45$	10

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