 Hello and welcome to the session. In this session we discussed the following question which says, the following table gives the daily income of 50 workers of a factory. In this table, we are given the daily income and the number of workers for each class interval of the daily income. We have to find the mean, mode and median of the above data. Before we move on to the solution, let's recall the formulae to find the mean, mode and median for a given data. First of all, we have mean that is x bar is equal to a plus summation f i u i upon summation f i this whole multiplied by h. Where we have this a is the assumed mean, h is the class size where f i is the frequency belonging to x i where x i is the class mark. Then we have u i is equal to x i minus a upon h where again x i is the class mark. Next we have the formula to find out the mode for a given data. Mode is equal to n plus f 1 minus f 0 upon 2f 1 minus f 0 minus f 2 this where multiplied by h. Where we have l is the lower limit of the modal class then h is the class size. f 1 is the frequency of the modal class is the frequency of the class preceding the modal class then is the frequency of the class exceeding the modal class. Then next we have the formula to find out the median this is equal to l plus n upon 2 minus cf this whole upon f and this whole multiplied by h. Where we have l is the lower limit as the median plus number of observations cf is the cumulative frequency of the class preceding the median class then f is the frequency of the median class which is the class size. Moving on to the solution now so we have this table in which we have the daily income that is the given class intervals number of workers that is the corresponding frequencies for the given class intervals f i x i which is the class mark then u i given by the formula x i minus a upon h f i u i and cf that is the cumulative frequency. Now x i that is the class mark is given by upper limit plus lower limit upon 2 so for the first class interval it would be 110 then for the next class interval it is 130 for the next class interval it is 150 for the next class interval it is 170 and for the next class interval it is 190. Now first of all we will find the assumed mean a we take the assumed mean a to be that x i which lies in the center of all x i's so from all x i's you can see that 150 lies in the center of the given x i's so we have taken a that is the assumed mean as 150. The class size h is equal to 20. Let us now find out u i for each class interval it is given by x i minus assumed mean a upon the class size h so for the first class interval it is 110 minus 150 upon 20 so it would be minus 2. For the next class interval it would be minus 1 for the next class interval it would be 0 for the interval 160 to 180 u i is equal to 1 for the interval 180 to 200 u i is 2. Now f i into u i that is f i multiplied by u i now 12 multiplied by minus 2 is minus 24 14 multiplied by minus 1 is minus 14 8 multiplied by 0 is 0 6 multiplied by 1 is 6 10 multiplied by 2 is 20. Now let's find out the cumulative frequencies for each class interval for the first class interval that is 100 to 120 cumulative frequency is given by 12. For the next class interval for the next class interval 120 to 140 it is given by 14 plus 12 that is 26 for the next class interval it is 8 plus 26 which is 34. Now for the next class interval it is 6 plus 34 that is 40 for the next class interval is 10 plus 40 that is 50. Now the total frequency that is submission f i is equal to 12 plus 14 plus 8 plus 6 plus 10 that comes out to be equal to 50 then submission f i u i is equal to minus 24 minus 14 plus 0 plus 6 plus 20 which comes out to be equal to minus 12. So we have submission f i u i is equal to minus 12. Now let's find out the mean first mean is given by x bar is equal to using this formula we will substitute the respective values and find out the mean. So we have mean x bar is equal to that is 150 plus submission f i u i which is minus 12 upon submission f i that is 50 and this whole multiplied by h which is the class size and that is 20. Now this 0 cancels with this 0 and we get this is equal to 150 minus 24 upon 5 which is equal to 750 minus 24 upon 5 that is we get this is equal to 726 upon 5 which comes out to be equal to 145.2. So we have the mean x bar is equal to 145.2. Next we will find out mode now before we find out mode we should know the modal class modal class is the class with the maximum frequency. Now from this table we find that the maximum frequency is 14 and so modal class is 120 to 140 is equal to putting the respective values in this formula we get mode is equal to L which is the lower limit of the modal class. And in this case it would be 120 so 120 plus f 1 which is the frequency of the modal class and that is 14 so here we have 14 minus f 0 which is the frequency of the class preceding the modal class. Now class preceding the modal class is 100 to 120 and its frequency is 12 so here we have 12 upon 2f 1 that is 2 into 14 minus f 0 which is the frequency of the class preceding the modal class and that is 12 minus f 2 which is the frequency of the class exceeding the modal class. And in this case the class exceeding the modal class is 140 to 160 and its frequency is 8 so here we have 8 and this whole multiplied by h which is the class size and that is 20. So this is equal to 120 plus 2 upon 28 minus 20 whole multiplied by 20 further equal to 120 plus 2 upon 8 multiplied by 20. Now 2 4 times is 8 and 4 5 times is 20 so this is equal to 120 plus 5 which is equal to 125 so we get mode is equal to 125. Next we will find the median now we have n is equal to 50 and n upon 2 would be equal to 50 upon 2 that is 25. First we will find the median class. Median class is the class whose cumulative frequency is greater than and nearest to and upon 2 and in this case median class would be the class whose cumulative frequency is greater than or nearest to 25. So look at this table now the cumulative frequency greater than 25 or nearest to 25 is 26 the class interval once 20 to 140 is the median class that is the class interval 120 to 140 is the median class. Now we will find out the median we will substitute the respective values in this formula to find out the median. So median is equal to l that is the lower limit of the median class and in this case it would be 120 plus n upon 2 which is 25 minus cf that is the cumulative frequency of the class preceding the median class. Now class preceding the median class is 100 to 120 its cumulative frequency is 12 so cf would be 12. So here we have 12 which is the frequency of the median class and that is 14 so here we have 14 and this whole multiplied by h that is the class size which is 20. So further we get this is equal to 120 plus 13 upon 14 into 20 now 2 7 times is 14 and 2 10 times is 20 and this is equal to 120 plus 130 upon 7. So further we get 840 plus 130 upon 7 that is we have 970 upon 7 which is equal to 138.57 so this is the median thus we have got the value for median also so we have got the values for mean, we have got the mode and median for the given data this completes the session hope you have understood the solution of this question.