 Hello and welcome to the session. In this session, we will study about measures of variation. Now measures of variation are used to describe distribution of the data and it also tells us about stretch of the data. Now we shall study inter-partial range and mean absolute deviation. First of all, let us discuss inter-partial range. Now suppose this data is given to us and here we have represented the data in the shape of a box. Now we know that medium divides the data in two halves. So here this vertical line in the middle of the data represents the medium since it is dividing the data in two equal halves. Now these halves are again divided into halves. Now here this vertical line in the middle of the low half of the given data divides the lower half of the given data in two halves. So this is called the lower quartile of the given data. Similarly this vertical line in the middle of upper half of the given data is called the upper quartile of the given data. Now here the lower quartile is denoted by q1, the medium or middle quartile is denoted by q2 and the upper quartile is denoted by q3. So quartiles divide the whole data set into four equal parts. This means here each part contains 25% of the data. Now here inter-partial range is the range of middle half of the data which means here inter-partial range starts from q1 to q3. So inter-partial range is equal to upper quartile minus lower quartile. This means inter-partial range is equal to q3 minus q1. Now if the inter-partial range is small then there is less variability and if inter-partial range is large then there is a greater variability. Now let us see one example for this and here for this data set we have to find the inter-partial range. Now let us start with this solution. First of all we will arrange the given data from the least to the greatest. So arranging we have 2, 2, 3, 3, 4, 5, 5, 6, 7, 9. Now here the number of terms in the given data is 10 which is even. Now for finding the inter-partial range first of all we have to find the median for the given data and then we will find the lower quartile and upper quartile of this data. Now here number of terms is even so median q2 is equal to the mean of middle two terms which is 4 plus 5 whole upon 2 which is equal to 9 upon 2 and this is 4.5. Now this median will lie in between 4 and 5. Now this is the lower half of the given data and this is the upper half of the given data. Now the median of lower half of the given data set is called the lower quartile and here the median of lower half is 3 so this is q1 and the median of upper half of data set is called the upper quartile. So here 6 is upper quartile. Now here inter-quartile range is equal to upper quartile q3 minus lower quartile q1 so this is equal to 6 minus 3 which is equal to 3. Now let us study what is mean absolute deviation. Now mean absolute deviation is equal to sum of absolute deviations from the mean upon number of terms in the data. This means it is the average of absolute values of the difference between the mean and each value in the data set and the difference between the values and mean is called deviations. And absolute deviation is the absolute value of the deviation which means it is the absolute value of the difference between mean each value in the data set. Now mean absolute deviation tells us about the spread of the data and if it is small then there is less variability and if it is large then there is a greater variability. Now let us discuss an example for this in which we have to find the mean absolute deviation of this data. First of all let us find the mean of the given data which is equal to the sum of the observations whole upon the number of observations which is 5. So this is equal to 15 upon 5 which is equal to 3. Now let us draw a table in which we will write these values and we will find the absolute deviation. So here we have drawn a table in the first column we have written the items in the second column we will find the deviations and in the third column we will find the absolute deviations. Now let us denote the items by x so deviations will be x minus mean and absolute deviations will be absolute value of x minus mean. Now mean is equal to 3. Now here when x is equal to 2 then deviation will be 2 minus 3 which is equal to minus 1 or x is equal to 2 again deviation will be minus 1. Then for x is equal to 3 deviation will be 3 minus 3 which is 0 then for 4 it will be 1 and again for 4 deviation will be 1. Now we will find absolute deviations now here when deviation is minus 1 absolute deviation is absolute value of minus 1 which is 1. Similarly we have got these absolute deviations now on adding these absolute deviations this is equal to 4. We know that mean absolute deviation is equal to sum of absolute deviations from mean upon number of terms in data. So here mean absolute deviation is equal to sum of absolute deviations from the mean which is 4 upon number of terms which is 5 so this is equal to 0.8. So here mean absolute deviation is 0.8 which is small it means the variation is less. Now let us discuss the effect of outlier. Now any striking deviation from the overall pattern means effects of outliers on mean and variance. Now outlier is a value in the data set which is either very large or small from all other values in the data set. Now let us see an example for this in which this data is given to us. First of all let us find the mean of the data which is equal to 16 upon 5 which is equal to 12. Now we have got mean is equal to 12 but no value in the data is anywhere near 12. Also when we calculate the mean absolute deviation for this data then that is equal to 14.8 which is very high. It means there is a greater variability but you can see that these values in the data do not vary much. So this high value of mean and mean absolute deviation is due to a very large value 39. It means 49 is the outlier for the given data. So from this example you can see the effect of an outlier. Now let us see how you can check an outlier in any given data. Now here consider this given data. Here you can see that number of terms is 7. So median will be the middle term and the lower quartile is 4 and upper quartile is 6. So here interquartile range is equal to 2. Now to check an outlier first of all we will multiply 1.5 with the interquartile range. So this is equal to 1.5 into 2 which is equal to 3. Then in the next step we will add this value to the upper quartile. So this is equal to 9 and we will subtract this value from the lower quartile. So this is 1. Now any value in the given data which is less than 1 or greater than 9 will be an outlier. Now in the data you can see that no numbers are less than 1 and 36 is greater than 9 so 36 is the outlier of the given data. And in this way we can represent this data in a box and with a plot. So in this session we have learnt about the measures of variation. And this completes our session. Hope you all have enjoyed this session.