The daily wages (in Rs.'00) of 9 workers in a small firm are: 20, 22, 25, 25, 25, 28, 30, 35, 40. Calculate Karl Pearson's coefficient of skewness and comment on its sign.

Question

Accepted Answer

**Data (in Rs.'00):** 20, 22, 25, 25, 25, 28, 30, 35, 40 ($N=9$).

**Step 1 — Mean.**
$$\bar{x}=\dfrac{20+22+25+25+25+28+30+35+40}{9}=\dfrac{250}{9}=27.78$$

**Step 2 — Mode.** The value 25 occurs three times, more often than any other value, so:
$$	ext{Mode}=25$$

**Step 3 — Standard deviation.** Deviations from the mean (27.78) and their squares:

| x | $x-\bar{x}$ | $(x-\bar{x})^2$ |
|---|---|---|
| 20 | −7.78 | 60.49 |
| 22 | −5.78 | 33.38 |
| 25 | −2.78 | 7.72 |
| 25 | −2.78 | 7.72 |
| 25 | −2.78 | 7.72 |
| 28 | 0.22 | 0.05 |
| 30 | 2.22 | 4.94 |
| 35 | 7.22 | 52.16 |
| 40 | 12.22 | 149.38 |

$$\Sigma(x-\bar{x})^2 = 323.56$$
$$\sigma=\sqrt{\dfrac{323.56}{9}}=\sqrt{35.95}=6.00$$

**Independent check of the standard deviation** using the direct formula $\sigma=\sqrt{\frac{\Sigma x^2}{N}-\bar{x}^2}$: $\Sigma x^2 = 400+484+625+625+625+784+900+1225+1600=7268$, so $\frac{7268}{9}-\left(\frac{250}{9}ight)^2 = 807.56-771.60=35.95$, giving $\sigma=6.00$ — matches Step 3 exactly. …

Q.The daily wages (in Rs.'00) of 9 workers in a small firm are: 20, 22, 25, 25, 25, 28, 30, 35, 40. Calculate Karl Pearson's coefficient of skewness and comment on its sign.

Concept understanding — Karl Pearson's Coefficient of Skewness

x	$x-\bar{x}$	$(x-\bar{x})^2$
20	−7.78	60.49
22	−5.78	33.38
25	−2.78	7.72
25	−2.78	7.72
25	−2.78	7.72
28	0.22	0.05
30	2.22	4.94
35	7.22	52.16
40	12.22	149.38

Q.The daily wages (in Rs.'00) of 9 workers in a small firm are: 20, 22, 25, 25, 25, 28, 30, 35, 40. Calculate Karl Pearson's coefficient of skewness and comment on its sign.

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