Calculate Karl Pearsonian’s coefficient of skewness Skp 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 |
Solution
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:
Skp = `("Mean" - "Mode")/"S.D."`
= `(45.75 - 53.6957)/(16.6914)`
= `-(7.9457)/(16.6914)`
∴ Skp = – 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
∴ Skp = `("Mean"-"Mode")/"S.D."`
= `(45.75 - 53.6957)/(16.6916)`
= `(-7.9457)/(16.6916)`
= – 0.4760
