 Hello and welcome to the session. In this session we will discuss the question which says that the temperatures in degree Celsius of two cities for latter half of the year are given to us, that is the temperatures for city A and city B are given to us and we have to compare and contrast measure of variation in two cities using interquadride range. Now before starting the solution of this question we should know a result. Now we know that quartiles divide the given data into quarters. Now here Q1 is the lower quartile, Q2 is the medium which is called the middle quartile and Q3 is called the upper quartile that is this vertical line is Q1, this vertical line is Q2 and this vertical line is Q3. So here the interquadride range starts from Q1 to Q3 therefore interquadride range is equal to Q3 minus Q1 or we can say interquadride range is equal to upper quartile minus lower quartile and this result will work out as a key idea for solving out the given question. Now let us start with the solution of the given question. Now in this question first of all we will find the interquadride range of both these series. First of all for city A let us arrange the data from least to the greatest and here arranging this data from the least to the greatest we have 11, 20, 30 to 48, 60 and 69. Now here the number of terms N in this given data is 1, 2, 3, 4, 5 and 6 which is even therefore the medium Q2 is the mean of middle two terms which is 32 plus 48 whole upon 2 which is equal to 80 upon 2 that is equal to 40. So medium is in between 32 and 48. Now the values below Q2 forms the lower half and the values above Q2 forms the upper half of the given data. Now medium of lower half is lower quartile and medium of upper half is upper quartile. Now in the lower half we have three terms which is odd. So middle value that is this value will be the medium of the lower half of the given data. So this is Q1 or you can say 20 is the lower quartile of the given data. The medium of upper half of the given data is which is the upper quartile of the given data or we can say Q3 is equal to 60. Now we have to find the interquartile range for this data. Now using this result which is given in the key idea interquartile range is equal to upper quartile Q3 minus lower quartile Q1 which is equal to 60 minus 20 which is equal to 40. Similarly we will find the interquartile range for CTB. Now for CTB arranging the data from the least to the greatest we have 20, 22, 32, 45, 56 and 60. Now here the number of terms is 6 which is even. So the medium Q2 is the mean of middle two terms which is equal to 32 plus 45 whole upon 2 which is equal to 77 upon 2 which is equal to 38.5. So the medium Q2 lies between 32 and 45. Now we know that medium of lower half is lower quartile and medium of upper half is upper quartile. Now in the lower half here we have three terms. So medium will be the middle term which is 22 therefore 22 is the lower quartile Q1 of the given data. Similarly 56 is the upper quartile Q3 of the given data. Now here the interquartile range for this data is equal to Q3 minus Q1 which is equal to 56 minus 22 which is equal to 34. Now interquartile range CTB is 34 for CTA which is you can see that medium for CTA is 40 and medium for CTB is 38.5. So here mediums for these cities are close for CTA is 40 and interquartile range for CTB is 34. Therefore we can say that data is more spread in CTA because quartile range is higher which shows data variability. So this is the conclusion of the given question and that's all for this session. Hope you all have enjoyed the session.