Calculate the standard deviation for the following continuous distribution by the actual mean method. | Class interval | 10–20 | 20–30 | 30–50 | 50–70 | 70–80 | | --- | --- | --- | --- | --- | --- | | Frequency (f) | 5 | 8 | 16 | 8 | 3 |

> [!TLDR] > Mean $= 40.5$, $\sum fd^2 = 11790$, so $\sigma = \sqrt{294.75} = 17.168$. **Step 1 — Mean.** $\sum fm = 75 + 200 + 640 + 480 + 225 = 1620$ and $\sum f = 40$, so $\bar{X} = \dfrac{1620}{40} = 40.5$. **Step 2 — Working table** (with $d = m - \bar{X}$). | C.I. | f | m | fm | d = m − 40.5 | fd | fd² | | --- | --- | --- | --- | --- | --- | --- | | 10–20 | 5 | 15 | 75 | −25.5 | −127.5 | 3251.25 | | 20–30 | 8 | 25 | 200 | −15.5 | −124.0 | 1922.00 | | 30–50 | 16 | 40 | 640 | −0.5 | −8.0 | 4.00 | | 50–70 | 8 | 60 | 480 | +19.5 | +156.0 | 3042.00 | …

Calculate the standard deviation for the following – Class 11 Economics

Worked Examples · Example 13

Calculate the standard deviation for the following continuous distribution by the actual mean method.

Class interval	10–20	20–30	30–50	50–70	70–80
Frequency (f)	5	8	16	8	3

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Mean $= 40.5$ , $\sum fd^2 = 11790$ , so $\sigma = \sqrt{294.75} = 17.168$ .

Step 1 — Mean. $\sum fm = 75 + 200 + 640 + 480 + 225 = 1620$ and $\sum f = 40$ , so $\bar{X} = \dfrac{1620}{40} = 40.5$ .

Step 2 — Working table (with $d = m - \bar{X}$ ).

C.I.	f	m	fm	d = m − 40.5	fd	fd²
10–20	5	15	75	−25.5	−127.5	3251.25
20–30	8	25	200	−15.5	−124.0	1922.00
30–50	16	40	640	−0.5	−8.0	4.00
50–70	8	60	480	+19.5	+156.0	3042.00

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