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101 karl pearson s coefficient of skewness according

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Unformatted text preview: y’ s coefficient of skewness (iii)Measure of skewness based on moments 7.10.1 Karl – Pearson’ s Coefficient of skewness: According to Karl – Pearson, the absolute measure of skewness = mean – mode. This measure is not suitable for making valid comparison of the skewness in two or more distributions because the unit of measurement may be different in different series. To avoid this difficulty use relative measure of skewness called Karl – Pearson’ s coefficient of skewness given by: Mean - Mode Karl – Pearson’ s Coefficient Skewness = S .D. In case of mode is ill – defined, the coefficient can be determined by the formula: 3(Mean - Median) Coefficient of skewness = S .D. Example 18: Calculate Karl – Pearson’ s coefficient of skewness for the following data. 25, 15, 23, 40, 27, 25, 23, 25, 20 175 Solution: Computation of Mean and Standard deviation : Short – cut method. Size Deviation from A=25 D 0 25 -10 15 -2 23 15 40 2 27 0 25 -2 23 0 25 -5 20 ∑d=-2 N=9 d2 0 100 4 225 4 0 4 0 25 ∑d2=362 ∑d n −2 = 25 + 9 = 25 – 0.22 = 24.78 Mean = A + = = σ ∑d2 ∑d − n n 362 −2 − 9 9 2 2 40.22 − 0.05 = 40.17 = 6.3 Mode = 25, as this size of item repeats 3 times Karl – Pearson’ s coefficient of skewness = = Mean - Mode S .D. 176 24.78 − 25 6.3 − 0.22 = 6.3 = − 0.03 = Example 19: Find the coefficient of skewness from the data given below Size : 3 45 6 7 8 9 10 Frequency: 7 10 14 35 102 136 43 8 Solution: Size Frequency (f) 3 4 5 6 7 8 9 10 7 10 14 35 102 136 43 8 N=355 Mean = A + = 6+ Deviation From A=6 (d) -3 -2 -1 0 1 2 3 4 ∑ fd N 480 355 = 6 + 1.35 = 7.35 Mode = 8 Coefficient of skewness = d2 fd fd2 9 4 1 0 1 4 9 16 -21 -20 -14 0 102 272 129 32 ∑fd=480 63 40 14 0 102 544 387 128 ∑fd2=1278 σ= = ∑ fd 2 ∑ fd − N N 1278 480 − 355 355 3.6 − 1.82 = 1.78 = 1.33 = Mean - Mode S .D. 177 2 2 = 7.35 - 8 1.33 = 0.65 1.33 = −0.5 Example 20: Find Karl – Pearson’ s coefficient of skewness for the given distribution: X : 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 F: 2 5 7 13 21 16 8 3 Solution: Mode lies in 20-25 group which contains the maximum frequency f...
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This note was uploaded on 01/18/2014 for the course BUS 100 taught by Professor Moshiri during the Winter '08 term at UC Riverside.

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