Compute the Variance and Standard Deviation of the following distribution using the step-deviation method:

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| No. of students | 6 | 10 | 18 | 11 | 5 |

Question

Compute the Variance and Standard Deviation of the following distribution using the step-deviation method:

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| No. of students | 6 | 10 | 18 | 11 | 5 |

Accepted Answer

**Step 1 — Set up mid-points, assumed mean and step-deviations.** Take assumed mean $A=25$ (the mid-point of the middle class 20–30) and class width $h=10$; compute $d=\dfrac{x-25}{10}$ for each class mid-point $x$.

| Marks | $f$ | Mid-point $x$ | $d=\frac{x-25}{10}$ | $fd$ | $fd^2$ |
|---|---|---|---|---|---|
| 0–10 | 6 | 5 | −2 | −12 | 24 |
| 10–20 | 10 | 15 | −1 | −10 | 10 |
| 20–30 | 18 | 25 | 0 | 0 | 0 |
| 30–40 | 11 | 35 | 1 | 11 | 11 |
| 40–50 | 5 | 45 | 2 | 10 | 20 |
| **Total** | **50** | | | **−1** | **65** |

**Step 2 — Compute the mean.**
$$\overline{x} = A + \dfrac{\sum fd}{N}	imes h = 25 + \dfrac{-1}{50}	imes 10 = 25 - 0.2 = 24.8$$

**Step 3 — Compute the Variance and Standard Deviation.**
$$\sigma^2 = h^2\left[\dfrac{\sum fd^2}{N}-\left(\dfrac{\sum fd}{N}ight)^2ight] = 100\left[\dfrac{65}{50}-\left(\dfrac{-1}{50}ight)^2ight] = 100\left[1.3-0.0004ight] = 129.96$$
$$\sigma = \sqrt{129.96} = 11.4$$ …

Marks	$f$	Mid-point $x$	$d=\frac{x-25}{10}$	$fd$	$fd^2$
0–10	6	5	−2	−12	24
10–20	10	15	−1	−10	10
20–30	18	25	0	0	0
30–40	11	35	1	11	11
40–50	5	45	2	10	20
Total	50			−1	65

🎓Unlock everything free for 14 days

Unlock everything free for 14 days