 Hi and welcome to the session. Today I'll help you with the following question which is a survey regarding the heights in centimeter of 50 girls of class 10th of a school was conducted and the following data was updated. Find the mean, median and mode of the above data. Before proceeding for the solution, let's see how to find out mean, median and mode. First of all mean is given by the formula sigma xi fi upon sigma fi where i goes from 1 to n. Here xi is the class mark and fi is the frequency. Now median is given by n plus n upon 2 minus cf upon f into h where n is the lower limit of median class n is the number of observations cf is the cumulative frequency of class preceding the median class f is the frequency of median class h is the class size assuming to be equal. And the median class is the class whose cumulative frequency is greater than and nearest to n upon 2. Now lastly the mode is given by the formula n plus f1 minus f0 upon 2f1 minus f0 minus f2 into h where n is the lower limit of the modal class h is size of the class interval all class sizes to be equal. f1 is the frequency of the modal class f0 is the frequency of the class preceding the modal class is the frequency of the class succeeding the modal class. Now modal class is the class with maximum frequency. This is the key idea for this question. Now let's see its solution. First of all we will find mean which is given by the formula sigma fi xi upon sigma fi. Now here we are already given in the question the height in centimeters and number of girls. So here number of girls is the frequency that is fi so here 50 is sigma fi. Now let's find the class mark. We know that class mark is given by upper class limit plus lower class limit upon 2. So here for the class interval 120 to 130 the class mark that is xi will be 120 plus 130 upon 2 that is 125. For this class interval the class mark that is xi will be 135 and so on. Next we want to find sigma fi xi so first of all let us find fi xi for this first of all we will multiply f1 with x1 that is 2 into 125 which is equal to 250. Similarly 8 into 135 is equal to 1080 12 into 145 is equal to 1740 20 into 155 is equal to 3100 and 8 into 165 is equal to 1320. Now adding all these we get sigma fi xi which is equal to 7,490. Now let us substitute the values over here to find out mean. Now mean is also given by x bar so here mean that is x bar will be equal to sigma fi xi that is 7,490 upon sigma fi that is 50. Which will be equal to 149.8 thus mean that is x bar is equal to 149.8. Now let us move on to medium which is given by the formula n plus n upon 2 minus cf upon f into h. For this first of all we need to find out the median class and we know that median class is the class whose cumulative frequency is greater than and here is 2n upon 2. So first of all here n is equal to 50 so n upon 2 will be equal to 50 upon 2 that is 25. Now let us find out the cumulative frequencies. So for this the cumulative frequency will be 2 then 2 plus 8 that is 10. Now 10 plus 12 that is 22 now 22 plus 20 that is 42 and lastly 42 plus 8 that is 50. Now the cumulative frequency which is greater than and here is 225 is 42. So the median class will be 152,160 that means we have median class as the class interval 152,160. Now here l is the lower limit of the median class which is 150 so l is 150 and upon 2 is 25. Cf is the cumulative frequency of class preceding the median class. So here this is the median class and the class preceding to median class is 140 to 150 and the cumulative frequency corresponding to this class is 22. So Cf is equal to 22, f is the frequency of median class which is 20, h is the class size which is 10 over here. Now let us substitute all these values in the formula to find out median. So here median will be equal to 150 plus 25 minus 22 upon 20 into 10 which will be equal to 150 plus 3 upon 2 that is 151.5. Thus median is equal to 151.5. Now let us find out mode which is given by the formula l plus f1 minus f0 upon 2f1 minus f0 minus f2 into h. Now here first of all we will find the modal class that is the class with maximum frequency. So here the maximum frequency is 20 and the class is 150 to 160 that means the modal class is 150 to 160. Now here also l is the lower limit of the modal class which is 150 so we have l equal to 150. f1 is the frequency of the modal class which is 20, f0 is the frequency of the class preceding the modal class that is 12 over here. f2 is the frequency of the class exceeding the modal class which is 8 and lastly h is the class size which is 10 over here. Now let us find out mode by substituting all these values. So mode will be equal to 150 plus 20 minus 12 upon 2 into 20 minus 12 minus 8 into 10. On simplifying this we will get 150 plus 8 upon 20 into 10 which will be equal to 154. Thus mode is equal to 154. Thus for the given data mean is equal to 149.8, median is equal to 151.5 and mode is equal to 154. With this we finish this session hope you must have understood the question goodbye take care and have a nice day.