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If dispersion measures amount of variation, then the direction of variation is measured by skewness. The most commonly used measure of skewness is Karl Pearson's measure given by the symbol Skp. It is a relative measure of skewness.

${S_{KP} = \frac{Mean-Mode}{Standard Deviation}}$

When the distribution is symmetrical then the value of coefficient of skewness is zero because the mean, median and mode coincide. If the co-efficient of skewness is a positive value then the distribution is positively skewed and when it is a negative value, then the distribution is negatively skewed. In terms of moments skewness is represented as follows:

${\beta_1 = \frac{\mu^2_3}{\mu^2_2} \\[7pt] \ Where\ \mu_3 = \frac{\sum(X- \bar X)^3}{N} \\[7pt] \, \mu_2 = \frac{\sum(X- \bar X)^2}{N}}$

If the value of ${\mu_3}$ is zero it implies symmetrical distribution. The higher the value of ${\mu_3}$, the greater is the symmetry. However ${\mu_3}$ do not tell us about the direction of skewness.

**Problem Statement:**

Information collected on the average strength of students of an IT course in two colleges is as follows:

Measure | College A | College B |
---|---|---|

Mean | 150 | 145 |

Median | 141 | 152 |

S.D | 30 | 30 |

Can we conclude that the two distributions are similar in their variation?

**Solution:**

A look at the information available reveals that both the colleges have equal dispersion of 30 students. However to establish if the two distributions are similar or not a more comprehensive analysis is required i.e. we need to work out a measure of skewness.

${S_{KP} = \frac{Mean-Mode}{Standard Deviation}}$

Value of mode is not given but it can be calculated by using the following formula:

${ Mode = 3 Median - 2 Mean \\[7pt]
College\ A: Mode = 3 (141) - 2 (150)\\[7pt]
\, = 423-300 = 123 \\[7pt]
S_{KP} = \frac{150-123}{30} \\[7pt]
\, = \frac{27}{30} = 0.9 \\[7pt]
\\[7pt]
College\ B: Mode = 3(152)-2 (145)\\[7pt]
\, = 456-290 \\[7pt]
\, S_kp = \frac{(142-166)}{30} \\[7pt]
\, = \frac{(-24)}{30} = -0.8 }$

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