 Hello and welcome to the session. In this session we discuss the following question which says the name of two samples of sizes 50 and 40 were respectively 63 and 54. The corresponding standard deviations were respectively 9 and 6, obtained the standard deviation of the combined sample. Before we move on to the solution, let's discuss the formula for the combined name and combined standard deviation. First we have the combined name given by x bar is equal to m1 x1 bar plus m2 x2 bar and this will be upon m1 plus m2. Then next formula is for the combined standard deviation given by sigma or sigma12 that is we have sigma is equal to square root of into sigma1 square plus d1 square this whole plus m2 into sigma2 square plus d2 square and this whole upon m1 plus m2. Now here we have considered two series like this be series1 is the as the series sigma1 is the under deviation as the series1 is equal to x1 bar minus x bar and we know that this x bar is the combined name for the series2 we have m2 of the series the static name of the series sigma2 is the under deviation of the series d2 is equal to x2 bar minus x bar this is the key idea that we would use for the solution of this question. Now consider the question in this we have given two samples their respective sizes their means and their corresponding standard deviations and we are supposed to find the standard deviation of the combined sample. So let's see the solution now. So here we have m1 which is the size of the first series which is given as 2 is the size of the second series which is 40. So we have m1 equal to 52 equal to when next x1 bar which is the arithmetic mean of the first series and it is given as 63 and the arithmetic mean of the second series is 54. So we have x1 bar equal to 63 and x2 bar equal to 54. So let us find out the combined mean which is denoted by x bar and this is equal to m1 x1 bar that is 50 into 63 plus m2 x2 bar which is 40 into 54 and this will upon m1 plus m2 that is 50 plus this is equal to 3150 plus 6 0 and this will upon 90 further this is equal to 5310 upon 90 this 0 cancels with 0 and 9. 50 9 times is 531 so we have x bar which is the combined mean is 50 without b1 which is equal to x1 bar minus x bar and so this is equal to now x1 bar is 63 so 63 minus 59 which is equal to that is d1 is equal to 4. Now next d2 is equal to x2 bar minus x bar now x2 bar is 59 which is equal to minus 5 that is d2 is equal to minus 5. Then we have the deviation given by sigma or sigma12 is equal to square root of m into sigma1 square plus d1 square the whole plus m2 into sigma2 square plus d2 square the whole and this will upon m1 plus m2. So now the combined standard deviation sigma12 is equal to square root of now we can substitute the values for n1 n2 sigma1 sigma2 and d1 d2 so here we have the standard deviation of first series which is 9 minus 9 sigma1 square would be 81 plus d1 square d1 is 4 4 square is 16 whole plus n2 which is 14 into 2 which is the standard deviation of the second series which is 6 so here 6 square would be 36 plus d2 square and d2 is minus 5 minus 5 square is 25 and this whole upon n1 plus n2 which is 50 plus 40. So here we have used sigma1 as 9 and sigma2 as so further this is equal to that is sigma12 is equal to square root of 50 into 97 plus 40 into 90 and this is further equal to 50 plus 40 this upon 90 this gives us sigma12 is equal to square root of 7290 upon 90 81 times is 729 so we have sigma12 is equal to square root of 81 which is equal to 9 therefore sigma12 is equal to 9 that is the combined standard deviation denoted by sigma12 or you can say sigma also is equal to 9 so this is our answer this completes the session hope you have understood the solution of this question.