 Hello friends. So in this session on statistics, we are going to deal with another method of finding arithmetic mean of a given data set Okay, so we are going to discuss something called step deviation method So in the previous two sessions, we saw direct method. We also saw Assumed mean method, which is also called shortcut method in this session. We are going to study step deviation method We'll explain with an example and then we'll also try to derive the formula of step deviation method Okay, now as I told you in the last class or last session where we discussed The assumed mean method the problem what the problem with the direct method was if the values of the variable is quite large Okay, and the frequency also is quite large Then what happens the calculation of xi fi that is if we multiply all the values with respective frequencies The multiplication becomes quite cumbersome. So hence to address that issue because in the you know earlier era, there was no computer or calculator so when statistics methods just statistical method were discovered or Formulated then there was no Calculator and computer so hence people derived easier methods of simplifying the process So that's what the objective is so in the previous Assumed mean method you saw we reduced the calculation load in step deviation method also we go one step further and Simplify calculation a little bit more. How do I do it? So as we found out di in the assumed mean method where we had to assume a Mean of the given data set and we in this example. We have assumed that 300 is the resume I told you last session also that you have to pick up a in such a way that you are you know The a lies somewhere in between the low lowest and the maximum values of the variables If you see the lowest possible value is 240 here and the highest is 350 So somewhat in the middle I have taken 300 you might argue that it is not the mean of 240 and 350 as well But then I have taken 300 purposefully because it is it will be easier for me to subtract 300 from various numbers So now if you see I subtracted 240 after subtracted 300 from 240 to get minus 60 So these are my di's I have calculated di's and now one step further is you assume another Value H which which is let's say 10 Okay, so we have to assume one value H such and what what is the purpose? So we're going to this divide all these di's by this value H And why are we doing it? And it will become a little while, you know after a little while that it is Simplifying my calculation. So whatever di's I found out I divided that further with H right, so I have purposefully chosen a Value 10 because division by 10 will be easier. You can also choose multiples of 20 hundreds depending upon the case. So in this case, I just want One digit decimal or let's say, you know numbers at unit space and that will be all So in this case what we have done is we have divided all the di's by H So that all these numbers are further reduced So if you can see minus 60 is reduced to minus 6 44 reduces to minus minus 44 reduces to minus 4.4 And so on and so forth now if you notice multiplying these you eyes with frequency values It must much easier than multiplication of these frequency values with either di's or excise themselves Isn't it? So hence now you see the calculation becomes much easier So if you see how I have calculated fi ui now it becomes much simpler And then I have some summed all these fi ui's to get minus 11.2 here now Once I get that what is the formula of X bar X bar is nothing but a that is assumed mean this is assumed mean assumed mean assumed mean and this H into Summation fi ui divided by summation fi Okay, so now if you deploy the values in this formula, you will again get 295 point 85 now the question arises. How do you arrive at this kind of a formula, right? So let us see now what is what was ui ui was given as X i minus a by H isn't it if you see here It's di ui is di by H and di is nothing but x i minus a so x i minus a upon H So hence guys, can I not write this as? ui H is equal to x i minus a or I can then say That what can I say about this? I can say X i is equal to a plus ui H Isn't it now what was X bar from direct method X bar was summation fi X i from i equals 1 to n divided by summation fi from again i is equal to 1 to n right So what is summation fi X i so let us replace X i by this value. Okay, so what can I get I will get I equals 1 to n fi stays as it is and X i is replaced by a plus ui times H and the entire thing is divided by summation fi i is equal to 1 to n Okay, now I told you that we can you know within the summation I can simplify it further So I is equal to 1 n. This is nothing but fi times a plus fi ui times H Isn't it fi ui times H divided by summation Fi from i is 1 to i to n Okay, now if there are two terms within the summation we can split the summation We learned this in the previous session itself. So summation This will be i is equal to 1 to n Fi a plus summation So these two terms with the plus sign Separated by the plus sign will be broken into two terms so one two n again, and this is fi ui H divided by summation i Is equal to 1 to n if I Okay, if you split these two terms the fraction then you will get summation fi a i equals 1 to n divided by summation fi i is equal to 1 to n plus plus what? summation i is equal to 1 to n fi ui H divided by summation i is equal to 1 to n fi Okay, so these are the rules of summation So if there are two summations and divide by same denominator, so I split the fraction now What is this term the left-hand side this term? What is this if you see a is constantly multiplied to all the fi So hence we learned in the last class as well that this a can be brought out of the summation because it's a constant so summation fi i is equal to 1 to n i is equal to 1 to n and then divided by summation i is equal to 1 to n fi and then this is plus Again here also H can be brought out so H is H is a constant so H is brought out after within you know before the summation sign so i is equal to 1 to n and this is fi Ui so all those constant terms which are being multiplied by multiplied with all the variables within the summation sign can be brought out Okay, so this is my final Term now what will happen if you see this I Summation fi and this is the same thing so hence it can be written as a Plus H is outside and then summation. What is the formula now? We arrived at 1 to n fi ui divided by summation fi i is equal to 1 To n isn't it if you see This is what the formula which you used to calculate the mean and the mean came out to be 295.85. This is the formula if you just Derived here is the derivation Okay, if you are wondering how did we operate on summation? So I'll just do a quick recap of rules of summation. So let us say if I have summation A x i where a is a constant and i is let's say varying from 1 to n. This is nothing but a x 1 plus a x 2 plus a x 3 plus so on and so forth till a x n Isn't it? This is what summation means now if you see you can pull a common and you can write a x 1 plus x 2 Plus dot dot till x n Is it not so hence a and this x 1 plus x 2 plus x n can be written as Summation i is equal to 1 to n x i Right, so hence summation a x i is nothing but a time summation x i so a can be brought out of the Summation term the other thing which you might be wondering how did it happen is let's say if i is equal to 1 to n And let us say we have x i plus y i Y I and this means what x 1 plus y 1 Plus x 2 plus y 2 Plus x 3 plus y 3 plus so on and so forth till x n plus y n So this can be written as x 1 plus x 2 plus so on and so forth till x n rearranging Plus y 1 plus y 2 plus so on and so forth till y n Which can be then written as summation x i from i is equal to 1 to n Plus summation y i from i is equal to 1 to n So this these two these two rules I explain because we use this in the above derivation So if you can see I had to split this Correct so if there is two terms within the summation so they can be broken down into two sums like this and Then I had used one more to pull out the a's for example here if you see this a I have pulled This a out this was a constant and similarly if h is there so I can pull the constant out of the submission which I just proved here Okay, so these two rules were used of you know to derive arrive at the final formula. Hope you understood the method. Thank you