 Hello and welcome to the session. Let us discuss the following question. Question says, in a return market, few vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of the boxes. This is the given table which represents the distribution of the mangoes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? First of all, let us understand step deviation method for finding the mean. According to step deviation method, mean is equal to a plus h multiplied by summation fiui upon summation fi where, now in this formula, x pi is the mean is assuming fi represents the frequency ui is equal to xi minus a upon h where xi is the class mark of the interval and h is the class size. And a is assuming, now we will use this formula as our key idea to solve the given question. Let us now start with the solution. Now first of all, we will rewrite the data given in the question. Now we are given this distribution of mangoes according to number of boxes. First of all, let us find out class mark for every class interval. Let us recall that class mark is equal to upper class limit plus lower class limit upon 2. Now in this interval, 52 is the upper class limit and 50 is the lower class limit. So 52 plus 50 upon 2 is equal to 51. Similarly, we can find class mark for this class interval. We know 55 plus 53 upon 2 is equal to 54. For this interval, class mark is 57 plus 56 upon 2 is equal to 57. Similarly, midpoint of this class interval is equal to 60 and midpoint of this class interval is equal to 63. Now we will choose one among these Xi's as assumed mean and denote it by A. Let it be 51. So we denote 51 by A. Now we will find the difference between A and each of the Xi's. That is, we will find the variation of assumed mean from each of the Xi's. So here we can write Xi minus A. Now clearly we can see X1 is equal to 51 and A is also equal to 51. Now we have to find X1 minus A. So it is equal to 51 minus 51. That is 0. So here we can write 0. Now clearly we can see X2 is equal to 54 is equal to 51. So the variation of assumed mean from X2 is equal to 54 minus 51, which is further equal to 3. So here we can write 3. Similarly, 57 minus 51 is equal to 6. Now here 63 minus 51 is equal to 9 and 63 minus 51 is equal to 12. Clearly we can see all these values in this column are multiples of 3. So if we divide the values in the entire column by 3, we would get smaller numbers to multiply by f i. We know f i represents the frequency and here frequency is represented by number of boxes. So we will denote it by f i. Now we will find U i by dividing all these values by 3. And we also know that 3 is the class size of each of the discrete class intervals. So here we can write U i is equal to xi minus A upon h and f is nothing but class size. So here we can write class size f is equal to 3. Now 0 divided by 3 is equal to 0. 3 divided by 3 is equal to 1. 6 divided by 3 is equal to 2. 9 divided by 3 is equal to 3 and 12 divided by 3 is equal to 4. Now we will find the product f i U i 15 multiplied by 0 is equal to 0. 110 multiplied by 1 is equal to 110. 135 multiplied by 2 is equal to 270. 115 multiplied by 3 is equal to 345 and 25 multiplied by 4 is equal to 100. Now we will find out summation f i. We know summation f i is equal to sum of all these frequencies. So sum of all these frequencies is equal to 400. Here we will find out summation f i U i. Summation f i U i is equal to sum of all these products. So sum of all these products is equal to 825. From key idea we know mean is equal to assume mean plus h multiplied by summation f i U i upon summation f i. Now substituting corresponding values of assume mean plus size summation f i U i and summation f i in this formula we get mean is equal to 51 plus 3 multiplied by 825 upon 400. Now we know 825 upon 400 is equal to 2.0625. So mean is equal to 51 plus 3 multiplied by 2.0625. Now multiplying these two terms we get 6.1875. So we can say mean is equal to 51 plus 6.1875. Now this implies mean is equal to 57.1875. Now rounding off this value up to two places of decimals we get mean is equal to 57.19. So we get the mean number of mangoes kept in a packing box is equal to 57.19. Now here we have used step deviation method for finding the mean. When the numerical values of x i and f i are large finding the product of x i and f i becomes tedious and time consuming. So for such situations we use step deviation method. So we can say we have used step deviation method because numerical values of x i and f i are large. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.