The following table gives the marks obtained by 50 students in a Statistics test. Calculate Karl Pearson's coefficient of skewness for this distribution.

| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| No. of students | 5 | 8 | 15 | 12 |

Question

The following table gives the marks obtained by 50 students in a Statistics test. Calculate Karl Pearson's coefficient of skewness for this distribution.

| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| No. of students | 5 | 8 | 15 | 12 | 7 | 3 |

Accepted Answer

**Step 1 — Arithmetic mean** (step-deviation method, assumed mean $A=35$, class width $h=10$):

| Marks | Mid-value $x$ | $f$ | $d=\frac{x-A}{h}$ | $fd$ | $fd^2$ |
|---|---|---|---|---|---|
| 0-10 | 5 | 5 | −3 | −15 | 45 |
| 10-20 | 15 | 8 | −2 | −16 | 32 |
| 20-30 | 25 | 15 | −1 | −15 | 15 |
| 30-40 | 35 | 12 | 0 | 0 | 0 |
| 40-50 | 45 | 7 | 1 | 7 | 7 |
| 50-60 | 55 | 3 | 2 | 6 | 12 |
| **Total** |  | **50** |  | **−33** | **111** |

$$\bar{x}=A+\dfrac{\Sigma fd}{N}	imes h = 35+\dfrac{-33}{50}	imes 10 = 35-6.6=28.4$$

**Step 2 — Mode.** The highest frequency (15) is in the class 20-30, so this is the modal class: $L=20,\ f_1=15,\ f_0=8,\ f_2=12,\ h=10$.
$$Z=L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}	imes h = 20+\dfrac{15-8}{30-8-12}	imes 10 = 20+\dfrac{7}{10}	imes 10 = 20+7=27$$

**Step 3 — Standard deviation.**
$$\sigma = h\sqrt{\dfrac{\Sigma fd^2}{N}-\left(\dfrac{\Sigma fd}{N}ight)^2} = 10\sqrt{\dfrac{111}{50}-\left(\dfrac{-33}{50}ight)^2} = 10\sqrt{2.22-0.4356} = 10\sqrt{1.7844}=13.36$$

**Step 4 — Karl Pearson's coefficient.**
$$Sk_P=\dfrac{\bar{x}-Z}{\sigma}=\dfrac{28.4-27}{13.36}=\dfrac{1.4}{13.36}=0.10$$ …

Marks	Mid-value $x$	$f$	$d=\frac{x-A}{h}$	$fd$	$fd^2$
0-10	5	5	−3	−15	45
10-20	15	8	−2	−16	32
20-30	25	15	−1	−15	15
30-40	35	12	0	0	0
40-50	45	7	1	7	7
50-60	55	3	2	6	12
Total		50		−33	111

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