Question:

1. Tabulate the data
| Class (students/teacher) | Frequency (f) | Mid‑point (x) |
|--------------------------|--------------|--------------|
| 15 – 20 | 3 | 17.5 |
| 20 – 25 | 8 | 22.5 |
| 25 – 30 | 9 | 27.5 |
| 30 – 35 | 10 | 32.5 |
| 35 – 40 | 3 | 37.5 |
| 40 – 45 | 0 | 42.5 |
| 45 – 50 | 0 | 47.5 |
| 50 – 55 | 2 | 52.5 |

2. Total frequency \(N = 3+8+9+10+3+0+0+2 = 35\).

3. Mean (grouped data)
Compute \(\sum f x\):
\(3\times17.5 = 52.5\)
\(8\times22.5 = 180\)
\(9\times27.5 = 247.5\)
\(10\times32.5 = 325\)
\(3\times37.5 = 112.5\)
\(2\times52.5 = 105\)
\(\sum f x = 52.5+180+247.5+325+112.5+105 = 1,022.5\)
\[\bar{x}=\frac{\sum f x}{N}=\frac{1,022.5}{35}=29.214\approx 29.2\]
Hence the mean teacher‑student ratio ≈ 29.2 students per teacher.

4. Mode (grouped data)
- Identify the modal class (largest frequency): 30 – 35 (\(f_1=10\)).
- Preceding class frequency \(f_0 = 9\) (25 – 30).
- Succeeding class frequency \(f_2 = 3\) (35 – 40).
- Class width \(h = 5\).
- Lower class boundary \(L = 30\).
- Apply the formula:
\[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h
=30+\frac{10-9}{2\times10-9-3}\times5
=30+\frac{1}{8}\times5
=30+0.625=30.625\]
Hence the mode ≈ 30.6 students per teacher.

5. Interpretation
- *Mode* (≈30.6) tells that the most frequently occurring teacher‑student ratio among the states lies in the 30‑35 interval; i.e., most states have about 31 students per teacher.
- *Mean* (≈29.2) gives the overall average ratio across all states. Since the mean is slightly lower than the mode, the distribution is mildly left‑skewed – a few states with lower ratios (e.g., the 15‑20 interval) pull the average down while the bulk of states cluster around the higher ratio indicated by the mode.
- Together, they indicate that while the typical state has around 31 students per teacher, on the whole India’s average is a little lower because of the presence of some states with comparatively better (lower) teacher‑student ratios.