 Hello and welcome to this session. In this session, we will discuss properties of arithmetic mean. Now we know that for given data, the arithmetic mean is the sum of observations over number of observations. Now the first property of arithmetic mean is the algebraic sum of the deviations of a set of values from the arithmetic mean is 0. Now let us start with its proof. Now let xi where i is equal to 1, 2 and so on up to n be the set of values and f be the corresponding frequencies of a frequency distribution. Then the algebraic sum of the deviations from the arithmetic mean is given by summation f into xi minus x bar the whole where i varies from 1 to n is equal to f1 into x1 minus x bar the whole plus f2 into x2 minus x bar the whole plus f3 into x3 minus x bar the whole plus 1 up to 1 into xn minus x bar the whole which is further equal to 1 x1 plus f2 x2 plus f3 x3 plus so on up to n the whole minus x bar into f1 plus f2 plus so on up to fn the whole. Now this expression can also be written as summation f y xi where i varies from 1 to n minus x bar into now this expression can be written as summation f5 where i varies from 1 to n. Now we know that the arithmetic mean of a given set of values that is x bar is equal to summation f i xi where i varies from 1 to n whole upon capital N. Precise capital N into x bar is equal to summation f i xi where i varies from 1 to n and also capital N is equal to summation f5 where i varies from 1 to n. So now putting this value to be equal to capital N into x bar minus now summation f5 where i varies from 1 to n is equal to capital N so this is equal to capital N into x bar which is further equal to 0. Hence the algebraic sum of the deviations of the set of values from their arithmetic mean is equal to 0. Hence we have proved the first property. Now let us start with the second property of the arithmetic mean and that is the sum of the squares of the deviations of the set of values is very much done taken about the mean. Now let us start with its proof. Now let xi over f i where i is equal to 1, 2, 3 and so on up to n be a frequency distribution that is for x1 observation f1 is the frequency for x2 observation f2 is the frequency and so on. Now let s is equal to summation f i into xi minus a whole square where i varies from 1 to n is the sum the squares of the deviations of given values from the arbitrary point. Now we have to prove that s is minimum when a is equal to x bar that is we have to prove that the sum of the squares of the deviations is minimum when taken about the mean where x bar is the mean. Now let this be equation number 1. Now in this equation on differentiating s with respect to a we get ds over da is equal to minus 2 into summation f i into xi minus a the whole where i varies from 1 to n. Now for maximum or minimum the derivative of s with respect to a will be equal to 0 that is ds over da is equal to 0 which implies minus 2 into summation f i into xi minus a the whole where i varies from 1 to n is equal to 0 which implies summation f i into xi minus a the whole where i varies from 1 to n is equal to 0 which further gives summation f i x i where i varies from 1 to n minus summation into a where i varies from 1 to n is equal to 0 which further implies summation f i x i where i varies from 1 to n is equal to summation f i into a, where i varies from 1 to n. Now this class summation f i x i where i varies from 1 to n is equal to a into summation f i where i varies from 1 to n. Now this further implies summation f i x i where i varies from 1 to n is equal to a into capital N because summation f i where i varies from 1 to n is equal to capital N. This further implies a is equal to 1 by capital N into summation f i x i where i varies from 1 to n. This gives a is equal to x bar because summation f i x i where i varies from 1 to n all upon capital N is equal to x bar that is the arithmetic mean. Now for d a is equal to 0 we have written a is equal to x bar, a is equal to x bar. Now for checking this we will find the second derivative of s with respect to a that is d square s over d a square and if d square s over d a square is less than 0 then s is maximum and if d square s over d a square is greater than 0 then s is minimum. Now this is d s over d f so d square s over d a square will be equal to 2 into summation f i where i varies from 1 to n which further implies d square s over d a square is equal to 2 into capital N because summation f i where i varies from 1 to n is equal to capital N which is that is 2 into n is greater than this that we are getting d square s over d a square greater than 0. Minimum a is equal to the deviations of a set of values is minimum when taken about the properties of arithmetic mean and this completes our session hope you all have enjoyed the session.