 Hello and welcome to the session. In this session we are going to discuss the following question which says that VHF 10-year slurs are noted as 210, 170, 187, 192, 190, 181, 150, 120, 200, 180 in kilograms from the above data find Q1, Q3 and interquartile range. VHF 10-year slurs are given as fellows and from this set of data we have to find lower quartile Q1, upper quartile Q3 and interquartile range. We know that for voting quartiles we first arrange the given series in ascending order. Then we have lower quartile Q1 given by n plus 1 divided by 4th item and upper quartile Q3 is equal to 3 by 4 n plus 1th item and interquartile range is equal to upper quartile Q3 minus lower quartile Q1 and n is the total number of elements in the series. We know that if n is the total number of elements in the series then lower quartile Q1 is given by n plus 1 whole divided by 4th item, upper quartile Q3 is given by 3 by 4 multiplied by n plus 1th item and interquartile range is given by the difference of upper quartile and lower quartile. With this key idea we proceed to the solution. Now in the question weights of 10-year slurs are given as fellows we have to calculate Q1, Q3 and interquartile range from this given set of data. Now using the key idea for getting quartiles we first need to arrange the given weights in ascending order that is first arrange the given weights ascending order then we get the ordered list of weights as 120, 150, 170, 180, 181, 187, 190, 192, 200 and 210. Here the weights of 10-year slurs are given so we get the value of n as 10. Now we first find Q1 that is lower quartile range using the key idea which gives the value of Q1 given by n plus 1 by 4th item. So we can write Q1 is equal to value of n plus 1 by 4th item. We know that the value of n is 10 so we can write value of 10 plus 1 by 4th item which is equal to the value of 2 decimal 7th 5th item which can be written as value of 2nd item plus decimal 7th 5th multiplied by value of 3rd minus 2nd item. Now from the series of weights we have the value of 2nd item as 150 and the value of 3rd item is 170. On tipping the values we get 150 plus decimal 7th 5th can be written as 3 by 4 multiplied by value of 3rd item which is 170 minus value of 2nd item which is 150 equal to 150 plus 3 by 4 multiplied by 20 which get cancelled with 4 and gives 150 plus 15 equal to 165. Hence we have got Q1 equal to 165. Now we shall find the value of upper quartile Q3 using the key idea which gives Q3 equal to 3 by 4 multiplied by n plus 1th item. So we write Q3 equal to value of 3 by 4 multiplied by n plus 1th item and value of n is 10 so we get value of 3 by 4 10 plus 1th item which further gives value of 8 decimal 2 5th item which can be written as value of 8th item plus decimal 2 5 multiplied by value of 9th item minus 8th item. That's the value of 8th item as 192 and the value of 9th item as 200. The value of 8th item is 192 plus decimal 2 5 can be written as 1 by 4 multiplied by value of 9th item which is 200 minus value of 8th item which is 192. On further simplification we get 192 plus 1 by 4 multiplied by 8 which is equal to 192 plus 2 which is equal to 194 so we get the value of Q3 equal to 192 plus 2 by 4 multiplied by 4. Now we shall find the value of Q3 equal to 192 plus 2 by 4 multiplied by Q3 equal to 192 plus 2 by 4 multiplied by Q3 equal to 192 and we have already calculated the value of Q1 and Q3. Now we can easily find the value of intercovertile drained using the key idea given by Q3 minus Q1. We know the value of Q3 is 194 minus the value of Q1 is 165 165 so we get the difference as 29 so we get the value of intercovertile drained as 29. Therefore we conclude that lower covertile Q1 is equal to 165 upper covertile Q3 is equal to 194 and intercovertile drained is equal to 29 which is our final answer. This completes the session. Hope you have understood it well.