 Hello and welcome to the session. In this session we will discuss weighted mean, mean of combined distributions and corrected mean. First we will discuss weighted mean. Now the weighted mean is similar to an arithmetic mean where instead of each of the data points, the distribution equally to final average, some data points contribute more than the others. Now, in simple arithmetic mean, each item has equal importance. Now let us discuss one example. For the arithmetic mean, now the marks of a student in four subjects 70 and 72. Therefore, the mean is equal to 40 plus 60 plus 70 plus 72. That is mean is equal to the sum of the observations over number of observations, 42 over 4 which is equal to 60.5. In simple arithmetic mean with importance, they are not of equal significance when weights are assigned to them according to their importance and weighted arithmetic mean is used as an average and the mean so obtained is called the weighted mean. Therefore, we can have that the weighted arithmetic mean when the values differ in their significance. Now let us discuss another example and that is in effective, 5 dollars at the wages then is equal to 2 plus 20 over 4 which is equal to 2.5 dollars. Different wages are considered to be of equal importance. If the number of wage earners are equal in all the cases, it is justified otherwise not. That is, let us say, we are earning 10 dollars and then 5 workers are earning 10 dollars or earning 20 dollars. Different wages are considered to be of equal importance in the arithmetic mean. The wages are given to us in dollars. That is, 20 and 20 and the corresponding number of wage earners are 6, 8, 10 and 5. That is, 6 workers are earning 5 dollars, 8 workers are earning 10 dollars, 10 workers are earning 20 dollars. Now here are the number of wage earners where each wage is different. Therefore, the weighted arithmetic mean of the wages is calculated as plus 10 into 8 plus 15 into 10 plus 20 into 5, 10 upon x plus 8 plus 10 plus 5 which is equal to 29 which is equal to 12.41. So we have got the weighted arithmetic mean as 12.41 dollars or we can say that we use weighted arithmetic mean when the values differ in their importance or significance. Now if the node of a variable x w1, w2, w3 and w1 denotes actively their weights, then their weighted mean xw bar is given as w1 x1 plus w2 x2 plus so on. Wn xn w1 plus w2 plus wn which is equal to wixi where i varies from 1 to n or upon summation wi where i varies from 1 to n or w bar that is the weighted mean is equal to summation wx over summation w. This is the formula for calculating the weighted mean. Now let us discuss the mean of the combined distributions. When two sets of scores have been combined into a single distribution, then the mean of the combined distribution is the weighted mean of the means of the components, the weights being the total frequencies in those components. Now in other words it is equal to n1 x1 bar plus n2 x2 bar over n1 is the mean of the combined distribution or the means distributions n1 and n2 are the total frequencies in distributions. Now let us start with express. Now x1 bar is equal to summation x over n1 or is equal to summation y over n2 is equal to n1 into x1 bar and this implies summation y is equal to n2 into x2 bar. Now the definition we have is equal to summation x plus summation y whole upon n1 plus n2 which is equal to now putting the values from here this will be n1 into x1 bar plus n2 into x2 bar whole upon n1 plus n2. Now in this formula the combines, now this formula may be extended to any number of distributions. Now if the combined distribution consists of three component parts then n1 is equal to n1 into x1 bar plus n2 into x2 bar plus n3 into x3 bar whole upon n1 plus n2 plus n3. And in general the two distributions x1 will be equal to n1 into x1 bar plus n2 into x2 bar plus so on up to nk into x3 bar and 1 plus n2 plus so on up to. This implies x bar is equal to summation ni into xi bar where i varies from 1 to k whole upon summation ni where i varies from 1 to k. Now let us discuss one example that is if the average salary of 80 workers was $600 that of 20 workers was $400 find the mean range of total number of workers. Now let us start with the solution now let that is the average salary of 80 workers is equal to and x2 bar that is the average salary of 20 workers is equal to $400 therefore we have n1 is equal to 80 and n2 is equal to 20. Therefore total number of workers that is x bar is equal to n1 into x1 bar plus n2 into x2 bar whole upon 100 plus 20 into 400 that is equal to 56,000 over 100 which is equal to for the mean range of total number of workers. So total number of workers due to mistake in copying certain items are only taken in finding the arithmetic mean in such cases we can calculate the new collected arithmetic mean by the procedure. Now in the first step multiply the incorrect arithmetic mean say x1 bar which is equal to summation arithmetic mean by n from the incorrect will be obtained to the value which we are obtaining in step 2 that is the correct summation of x will be equal to. Now in the last step we can find the correct arithmetic mean that is the correct summation of x which we are obtaining in step 3 over the number of observations that is an example in this 75 but later it was discovered that 13 was miscopied as 31 find the correct mean. Arithmetic mean is 75 that is incorrect x bar is equal to 75 and the number of observations is the number of observations which further implies summation x incorrect arithmetic mean which is equal to 15 into 75 which is equal to 1,125 is copied as 31. The correct sum of observations that is the correct summation of x by using this formula is incorrect minus the correct item. Suppose this copied as 31 this means this is equal to 0,25 minus 13 into 107 mean is equal to the correct sum of observations that is 1,107 over the number of observations that is 15 which is equal to 73.8 and find the correct mean. So in this session we have learnt about related mean, mean of combined distributions. Hope you all have enjoyed the session.