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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 - Mathematics and Statistics

Sum

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
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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

Concept: Measures of Skewness - Karl Pearson’S Coefficient of Skewness (Pearsonian Coefficient of Skewness)
  Is there an error in this question or solution?
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