 Hello and welcome to the session. In this session we will discuss a question which says that a fly vendor keeps the records of a sales in dollars for 25 days. He records for following observations and these are the 25 observations. And here we have to make a box plot for the data and identify the outliers. Now let us start with the solution of the given question. Now here we have to make a box plot for this given data set. So we should find median into the child range for this data set. Now arranged in the given data set in ascending order we have 15, 15, 15, 15 and again 15. Then we have 100, 100, 100, 100, then 150, 200, 200, then 250, 250. Next we have 300, 350, 450, then we have 700, 1000, 1650, 1700 and lastly we have 2750. Now first of all let us find the median for this given data. Now here number of terms that is n is equal to 25 which is odd. So median is the middle value which is equal to 13th term of the given data. Now here we can see that 250 is the 13th term of this data. So median is equal to 250. Now 12 terms or we can say 12 items lie below the median and 12 items lie above the median. Now for lower quartile we will find the median of the lower 12 terms. Now median is the middle value. There are 12 terms that is even number of terms. So middle value that is the lower quartile will be mean of middle three terms which is equal to mean of seventh term. So lower quartile let us denote it by q1 is equal to mean of sixth and seventh term that is 100 plus 100 whole upon 2 which is equal to 200 upon 2 and this is equal to 100. Now upper quartile is the middle value of the upper 12 terms of this data and this is equal to mean of 19th and 20th term of this data. So upper quartile as is denoted by q3 is equal to 450 plus 500 and this is equal to 950 upon 2 which is equal to 475. This is equal to upper quartile minus lower quartile that is q3 minus q1 which is equal to 475 minus 100 and this is equal to 375. And now let us check for the outliers. For us we see the range q1 minus 1.5 into interquartile range q3 plus 1.5 into interquartile range. First of all let us find q1 minus 1.5 into interquartile range. Now we know that lower quartile q1 is 100 minus 1.5 into interquartile range is 375 and this is equal to 100 minus 562.5 which is equal to minus 462.5. Now let us find this value. Now upper quartile q3 is 475 plus 1.5 into interquartile range is 375 which is equal to 475 plus 562.5 and this is equal to 1037.5 which is equal to minus 462.5 and above 1037.5 that is the values of the data search lying below minus above 1000 are outliers. Now in this data you can see there are no values below minus 462.5 but above 1037.5 we have 1650, 1700 and 2750. So these are the outliers. Now using all these values let us draw box dot. Now where we have taken? Scale of 500. So we draw the box with viscose and here the box starts from the lower quartile q1 to the upper quartile q3 and this think vertical line represents the median. Now here we have extended the lower viscose up to 0 because 0 is the lowest value of the given data and the upper viscose is extended up to 1000 and we do not cover the outliers in viscose. Here these three points represent outliers. This is the required box plot for the given data and this completes our session. Hope you all have enjoyed the session.