The daily cash sales (in ₹ thousand) of a shopkeeper over 6 days are: 20, 24, 18, 26, 22, 28. Find the Variance and Standard Deviation (a) by the direct method, and (b) by the assumed-mean method (take $A=22$), and verify the two agree.

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Accepted Answer

**Step 1 — Direct method: find the mean.** $\sum x=20+24+18+26+22+28=138$, $n=6$, $\bar{x}=138/6=23$.

**Step 2 — Direct method: squared deviations from the actual mean.**

| $x$ | 20 | 24 | 18 | 26 | 22 | 28 |
|---|---|---|---|---|---|---|
| $x-23$ | $-3$ | 1 | $-5$ | 3 | $-1$ | 5 |
| $(x-23)^2$ | 9 | 1 | 25 | 9 | 1 | 25 |

$\sum(x-\bar{x})^2 = 9+1+25+9+1+25 = 70$.
$$\sigma^2 = \dfrac{70}{6} \approx 11.6667, \qquad \sigma = \sqrt{11.6667} \approx 3.4157$$

**Step 3 — Cross-check by the assumed-mean method, $A=22$.**

| $x$ | 20 | 24 | 18 | 26 | 22 | 28 |
|---|---|---|---|---|---|---|
| $d=x-22$ | $-2$ | 2 | $-4$ | 4 | 0 | 6 |
| $d^2$ | 4 | 4 | 16 | 16 | 0 | 36 |

$\sum d = -2+2-4+4+0+6 = 6$; $\sum d^2 = 4+4+16+16+0+36 = 76$.
$$\sigma^2 = \dfrac{\sum d^2}{n} - \left(\dfrac{\sum d}{n}\right)^2 = \dfrac{76}{6} - \left(\dfrac{6}{6}\right)^2 = 12.6667 - 1 = 11.6667$$ …

Q.The daily cash sales (in ₹ thousand) of a shopkeeper over 6 days are: 20, 24, 18, 26, 22, 28. Find the Variance and Standard Deviation

Concept understanding — Standard Deviation and Variance

Q.The daily cash sales (in ₹ thousand) of a shopkeeper over 6 days are: 20, 24, 18, 26, 22, 28. Find the Variance and Standard Deviation

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