 Hello and welcome to the session. Let us discuss the following question. Question says, the following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency F. This is the given distribution table. First of all, let us understand that mean is equal to summation F i X i upon summation F i. Where X bar is the mean, F i represents the frequency and X i represents the midpoint of the interval or we can say it represents class mark of the interval. This is the key idea to solve the given question. Let us now start with the solution. First of all, we will rewrite the data given in the question. Now we are given daily pocket allowance in rupees and number of children. Now we know number of children represents the frequency and it is denoted by F i. Now let us find out midpoint of every class interval or we can say we will find out class mark of every class interval. We represent class mark by X i. Now upper limit plus lower limit upon 2 is equal to class mark. So for this interval, 13 is the upper limit, 11 is the lower limit and 13 plus 11 upon 2 is equal to 12. So here we will write 12. Similarly, midpoint of this interval is equal to 14. Now class mark of this interval is equal to 16. Here also 17 plus 19 upon 2 is equal to 18. Similarly, 19 plus 21 upon 2 is equal to 20 and 21 plus 23 upon 2 is equal to 22 and midpoint of this interval is equal to 24. Now we will find the product F i X i. Now 12 multiplied by 7 is equal to 84 or we can say 7 multiplied by 12 is equal to 84. Similarly, 6 multiplied by 14 is equal to 84. 9 multiplied by 16 is equal to 144. 13 multiplied by 18 is equal to 234. F multiplied by 20 is equal to 20 F. 5 multiplied by 22 is equal to 110 and 4 multiplied by 24 is equal to 96. Now let us find out summation F i. We know summation F i is equal to sum of all these frequencies and sum of all these frequencies is equal to 44 plus F. We know this frequency is unknown and sum of all these frequencies is equal to 44. So summation F i is equal to 44 plus F. Now let us find out summation F i X i. We know summation F i X i is equal to sum of all these products and sum of all these products is equal to 20 F plus 752. Adding all these light terms we get 752 and total sum of the products is equal to 20 F plus 752. Now in the question we are given that mean pocket allowance is equal to rupees 18 and from key idea we also know that mean is equal to summation F i X i upon summation F i. This is the direct method of finding the mean. We are given mean pocket allowance or we can say x bar is equal to rupees 18. Now substituting corresponding values of x bar summation F i and summation F i X i in this formula we get 18 is equal to 20 F plus 752 upon 44 plus F. Now multiplying both the sides of this expression by 44 plus F we get 18 multiplied by 44 plus F is equal to 20 F plus 752. Now we know 18 multiplied by 44 is equal to 792 and 18 multiplied by F is equal to 18 F. So we get 792 plus 18 F is equal to 20 F plus 752. Now subtracting 752 from both the sides we get 792 minus 752 plus 18 F is equal to 20 F. Now subtracting 18 F from both the sides we get 792 minus 752 is equal to 20 F minus 18 F. 792 minus 752 is equal to 40 and 20 F minus 18 F is equal to 2 F. So we get 40 is equal to 2 F. Now dividing both the sides of this expression by 2 we get 20 is equal to F or we can simply write it as F is equal to 20. So we get missing frequency F is equal to 20. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.