 Hi and welcome to the session. Let us discuss the following question. The question says, Find the mean deviation about the mean for the data in exercises 9 and 10. This is the data given to us in question number 9. We have to find the mean deviation about the mean of this data. Let's now learn the steps which are involved in the calculation of mean deviation about the of a continuous frequency distribution. In the first step, we have to obtain the midpoint of each class interval which is denoted by xi and this xi is equal to upper limit plus lower limit divided by 2 and then we have to find the mean of the given data by using the formula x1 is equal to 1 by n into summation i goes from 1 to n pi xi where n is equal to summation i goes from 1 to n pi. In the second step, we have to find the deviation of each xi from x bar that is x1 minus x bar x2 minus x bar so on xn minus x bar. In the third step, we have to find the absolute value of each deviation that is drop the minus sign if it is there that is mod of x1 minus x bar mod of x2 minus x bar so on mod of xn minus x bar and then in the last step, we have to find the mean of the absolute values of the deviation. This mean is the mean deviation about mean that is mean deviation about mean is equal to 1 by n into summation i goes from 1 to n pi into mod of xi minus x bar where n is equal to summation i goes from 1 to n pi. Always remember these steps so keeping all these steps in mind let's now begin with the solution. We will first make a table. We have written the given information in the first two columns. Now we will first find midpoint of each class interval and we know that midpoint is denoted by xi. We know that midpoint is obtained by adding upper limit and lower limit and then dividing this by 2. So using this midpoint of class interval 0 to 100 is 0 plus 100 divided by 2 and this is equal to 50. Midpoint of this class interval is 100 plus 200 by 2 and this is equal to 300 by 2 and 300 by 2 is 150. Then we have 200 plus 300 divided by 2 and this is 250. Then we have 300 plus 400 divided by 2 which is equal to 350. Midpoint of this class interval is 400 plus 500 by 2 and this is equal to 450. Then we have 500 plus 600 divided by 2 and this is equal to 550. Then we have 600 plus 700 by 2 and this is equal to 650 and then we have 700 plus 800 by 2 and this is equal to 750. Product of xi and fi. In the first row fi is 4 and xi is 50 so we are 4 into 50 and 4 into 50 is 200. 8 into 150 is 1200. 9 into 250 is 2250. 10 into 350 is 3500. 7 into 450 is 3150. 5 into 550 is 2750. 4 into 650 is 2600. 3 into 750 is 2250. Now we will find and we know that n is equal to summation i goes from 1 to n. Now here n is equal to 8 so we have summation i goes from 1 to 8 fi and this is equal to 4 plus 8 plus 9 plus 10 plus 7 plus 5 plus 4 plus 3 and this is equal to 50. So on adding all these frequencies we get 50. Now we will find summation i goes from 1 to 8 fi xi. Now this is equal to 200 plus 1200 plus 2250 plus 3500 plus 3150 plus 2750 plus 2600 plus 2250 and this is equal to 17900. We know that mean is equal to 1 by n into summation i goes from 1 to 8 fi xi. Substituting values of n and summation i goes from 1 to 8 fi xi we get 1 by 50 into 17900 and this is equal to 350a. So mean of the given data is 358. Now we are going to find xi minus x path. In the first row xi is 50 and we know that x path is equal to 358 so we have 15 minus 358 and 50 minus 358 is equal to minus 308. Then we have 150 minus 358 and 150 minus 358 is equal to minus 208. 250 minus 358 is minus 408. 350 minus 358 is minus 8. 450 minus 358 is 92. 50 minus 358 is 192. Fixed 50 minus 358 is 292. 750 minus 358 is 392. Now we will find mod of xi minus x path. Absolute value of minus 308 is 308. Absolute value of minus 208 is 208. Absolute value of minus 108 is 108. Absolute value of minus 8 is 8 and absolute value of 92, 192, 292 and 392 is 92, 192, 292 and 392. Find fi into mod xi minus x path. In the first row fi is 4 and mod xi minus x path is 308. So we have 4 into 308 and 4 into 308 is 1232. 8 into 208 is 1664. 9 into 108 is 972. N into 8 is 87 into 92 is 654. 5 into 192 is 960. 4 into 292 is 1168. 3 into 392 is 1176. When adding all this, we get 7896. So summation i goes from 1 to 8 fi into mod xi minus x path is equal to 7896. Now we will calculate mean deviation about mean. We know that mean deviation about mean is equal to 1 by n into summation. i goes from 1 to 8 fi into mod xi minus x path and this is equal to 1 by 50 into 7896 and this is equal to 157.92. Hence our required answer is 157.92. So this completes this action by intake care.