 Hello students, let's work out the following problem. It says the arithmetic mean of the following frequency distribution is 50 find the value of P. So this is the given frequency distribution and we have to find the value of P. So let's now move on to the solution. The mean of the frequency distribution is given by the formula summation f i x i upon summation f i where f i is the frequency x i is the midpoint of the class interval also called as class mark and f i is the frequency as you know. So this is the frequency distribution table. Now we have to find summation f i and summation f i x i. So we first need to find x i x i is the midpoint of the class interval. So it is 0 plus 20 divided by 2 that is 10 and this is 20 plus 40 divided by 2 that is 30 40 plus 60 divided by 2 is 50 60 plus 80 divided by 2 is 70 80 plus 100 divided by 2 is 90. Now 10 into 17 is 170 13 into P is 30 P 50 into 32 is 1600 17 into 24 is 1600 80 90 into 19 is 1700 10. Now we find the sum of f i into x i that is summation f i x i. Now 170 plus 1600 plus 1600 80 plus 1700 is equal to 5160 plus 30 P. Similarly we find summation f i now this is 17 plus 32 plus 24 plus 19 is 92 plus P mean is 50 which is given by the formula summation f i x i upon summation f i that is 5160 plus 30 P divided by 92 plus P. So this implies 50 into 92 plus P is equal to 5160 plus 30 P. Now again 15 to 92 is 4600 plus 15 to P is 50 P is equal to 5160 plus 30 P. Now this implies 4600 minus 5160 is equal to 30 P minus 50 P. Now 4600 minus 5160 is 560 it's minus 560 minus 20 P. So this implies 560 is equal to 20 P cancelling negative sign from both sides. So this implies P is equal to 560 upon 20 so this implies P is equal to 56 upon 2 that is equal to 28. Hence value of P is 28. So this completes the question and the session. Bye for now. Take care. Have a good one.