Statistics : Skewing

Statistics : Skewing

Normal Distribution

Symmetrical

For a symmetrical distribution the mean and the median are approximately in the middle and the upper and lower quartiles placed on either sides of the median. When the bulk of the data leans toward one or the other end of the range, the distribution it is said to be skewed.

Skewing in Box & Whiskers

Positively Skewed.

* The right side whisker is longer due to the presence of outliers. ( a few high values)

* The bulk of the data is in the lower end.

* The Median is closer to the Q1 quartile.

Negatively Skewed

*The left side whisker is longer due to the presence of outliers (small values)

* The median is closer to the right side / upper quartile.

* The bulk of the data is on the higher end or values.

Comparing 2 Box and Whisker Plot

* B - The data is close to being symmetrical / slightly symmetrical.

* A - The data is negatively skewed.

* IQR of A is less than the IQR of B.

*The bulk of the data in A had less spread than the bulk of the data in B.

* IQR of A is more constricted and IQR of B is more spread out.

Remember : IQR (inter- quartile range ) = Q3 - Q1

Standard Deviation

It is another measure of spread about the mean. I.Q.R. measures spread about the MEDIAN. The bulk of the data (for a normal distribution ) is about 2 Standard Deviation on either side of the MEAN.

1. Discrete Data

 $S.D.=\sqrt{\frac{\in\left(x-\overline{x}\right)^2}{n}}$   where

  $\overline{x}\ =mean\$  and x = individual (marks etc) 
n = number of items / raw data

eg. 1

Find the Standard Deviation of the following marks (Test out of 20) for Class A.
3 , 5 ,11 , 8 , 4 ,15, 17, 19, 6, 9. 
n = 10 

 $Mean\ =\frac{\in x}{n}$  
 $=\frac{97}{10}$  
 $=9.7$  

 $S.D.\ =\sqrt{\frac{\in\left(x-\overline{x}\right)^2}{n}\ }=\ \sqrt{\frac{286.1}{10}}=5.34$  

e.g 2

Find the S.D of the following set of data (Test out of 20) for Class B.
10 , 9, 15, 16, 17, 18, 20, 19, 11, 15

 $Mean\ =\frac{\in x}{n}=\frac{150}{10}=15$  

 $S.D.=\sqrt{\frac{\in\left(x-\overline{x}\right)^2}{n}\ }=\sqrt{\frac{132}{10}}=3.63$  
Class B has a smaller S.D. which means that there were less Spread of marks about the mean.