Question

Calculate Karl Pearsonian’s coefficient of skewness Sk_p from the following data:

Marks above	0	10	20	30	40	50	60	70	80
No. of students	120	115	108	98	85	60	18	5	0

Answer 1

The given table is the cumulative frequency table of more than type. From this table, we have to prepare the frequency distribution table and then calculate the value of Skp.
Construct the following table:

Mark above	No. of students 'more than'  (c.f.)	Class-interval	Frequency fi	Mid value xi	fixi	fixi2
0	120	0 – 10	5	5	25	125
10	115	10 – 20	7	15	105	1575
20	108	20 – 30	10	25	250	6250
30	98	30 – 40	13	35	455	15925
40	85	40 – 50	25	45	1125	50625
50	60	50 – 60	42	55	2310	127050
60	18	60 – 70	13	65	845	54925
70	5	70 – 80	5	75	375	28125
80	0	80 – 90	0	85	0	0
		Total	120	–	5490	284600

From the table, N = 120, `sumf_"i"x_"i" = 5490   "and" sumf_"i"x_"i"^2 = 284600`

Mean = `bar(x) - (sumf_"i"x_"i")/"N" = 5490/120` = 45.75

Maximum frequency 42 is of the class 50 – 60.
∴ Mode lies in the class 50 – 60.
∴ L = 50, f1 = 42, f0 = 25, f2 = 13, h = 10

∴ Mode = `"L" + ("f"_1 - "f"_0)/(2"f"_1 - "f"_0 - "f"_2) xx "h"`

= `50 + (42 - 25)/(2(42) - 25 - 13) xx 10`

= `50 + 17/(84 - 38) xx 10`

= `50 + 17/46 xx 10`

= 50 + 3.6957
= 53.6957

S.D. = `sqrt((sumf_"i"x_"i"^2)/"N" - (bar(x))^2`

= `sqrt(284600/120 - (45.75)^2`

= `sqrt(2371.6667 - 2093.0625)`
= `sqrt(278.6042)`
= 16.6914
Pearsonian’s coefficient of skewness:

Sk_p = `("Mean" - "Mode")/"S.D."`

= `(45.75 - 53.6957)/(16.6914)`

= `-(7.9457)/(16.6914)`

∴ Sk_p = – 0.4760
Alternate Method:

Let u = `(x - 45)/10`

Mark above	No. of students ‘more than’ (c.f.)	Class	Frequency (fi)	Mid value xi	ui	fiui	fiui2
0	120	0 –10	5	5	− 4	− 20	80
10	115	10 –20	7	15	− 3	− 21	63
20	108	20 – 30	10	25	− 2	− 20	40
30	98	30 –40	13	35	− 1	− 13	13
40	85	40 – 50	25	45	0	0	0
50	60	50 –60	42	55	1	42	42
60	18	60 –70	13	65	2	26	52
70	5	70 –80	5	75	3	15	45
80	0	80 –90	0	85	4	0	0
		Total	120			9	335

`bar(u) = (sumf_"i"u_"i")/"N" = 9/120` = 0.075

∴ `bar(x) = 45 + 10(bar(u))`

= 45 + 10(0.075)
= 45 + 0.75
= 45.75

Var(u) = `sigma_u^2 = (sumf_"i"u_"i"^2)/"N" - (bar(u))^2`

= `335/120 - (0.075)^2`

= 2.7917 – 0.0056
= 2.7861
Var(X) = h² x Var(u) = 100 × 2.7861 = 278.61
S.D. = `sqrt(278.61)`
= 16.6916
Maximum frequency 42 is of the class 50 – 60.
∴ Mode lies in the class 50 – 60.
∴ L = 50, f1 = 42, f0 = 25, f2 = 13, h = 10

∴ Mode = `"L" + ("f"_1 - "f"_0)/(2"f"_1 - "f"_0 - "f"_2) xx "h"`

= `50 + (42 - 25)/(2(42) - 25 - 13) xx 10`

= `50 + 17/(84 - 38) xx 10`

= `50 + 17/46 xx 10`

= 50 + 3.6957
= 53.6957

∴ Sk_p = `("Mean"-"Mode")/"S.D."`

= `(45.75 - 53.6957)/(16.6916)`

= `(-7.9457)/(16.6916)`

= – 0.4760