# Is this method to find mean already discovered?

I am a 10th class student and in our syllabus, we have three methods for finding mean of grouped data:

• Direct method.
• Assumed mean method.
• Step deviation method.

Out of these, the Step deviation method is the simplest but still requires a lot of calculations. In the step-deviation method, you have to first find class mark ($x_i$), subtract some number ($a$) from all of them to get $d_i$ and then divide them all by some number ($h$) to get $u_i$, and then use a formula to get mean. After doing some of the excercises, I noticed that in most of the questions, the values of ui were ...,-2,-1,0,1,2,... etc., so I made this method.

Suppose this is the data:

$$\begin{array}{c|cc} i & \text{CI} & f_i \\\hline 1 & 1-3 & 1\\ 2 & 3-5 & 2\\ 3 & 5-7 & 2\\ 4 & 7-9 & 1\\ \end{array}$$

1. Let $m=$ the class number with the largest $f_i$. (It doesn't matter which number you choose but it will be easy in this way). $\qquad\therefore \qquad m=3$.

2. Set $k_i = m-i$. $$\begin{array}{c|cc|c} i & \text{CI} & f_i & k_i \\\hline 1 & 1-3 & 1 & 2\\ 2 & 3-5 & 2 & 1\\ 3 & 5-7 & 2 & 0\\ 4 & 7-9 & 1 & -1\\ \end{array}$$

3. Let Mean of $kf = \bar k$, that is, $\bar k = \frac{\displaystyle \Sigma f_ik_i}{\displaystyle \Sigma f_i }$ $$\begin{array}{c|cc|c|c} i & \text{CI} & f_i & k_i&f_ik_i \\\hline 1 & 1-3 & 1 & 2 & 2\\ 2 & 3-5 & 2 & 1 & 2\\ 3 & 5-7 & 2 & 0 & 0\\ 4 & 7-9 & 1 & -1&-1\\\hline &&\Sigma f_i = 6&&\Sigma f_ik_i=3 \end{array}$$$$\therefore\qquad\bar k=\frac12$$

4. Use this formula which I discovered: $$\bar x = -h\bar k+l+hm-\frac h2$$ (where $h$ is class size and $l$ is lower limit of the first class) $$\therefore\qquad\bar x = -2\times\frac12+1+2\times 3-\frac22=5$$

Which is the correct answer.

Now the question is

1. Has this method been dicovered earlier? What do we call it?
2. If it has been dicovered earlier, then why do they still teach us so complicated methods in school?

(As we all know, math is ungooglable.)

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How do you define $u_i$? –  dani_s Apr 19 '14 at 14:35
Whoever downvoted, please at least give a reason? –  Kartik Apr 20 '14 at 14:39
Not sure who downvoted. I deleted my answer since it was no longer relevant after you made your edit. I haven't really looked at all these different methods, which are optimized for hand calculation, and therefore largely obsolete. From your description, it seems that the step deviation method should work even when the class sizes are unequal. If you can assume all of the class sizes to be the same, then your method is indeed a simplification. –  Will Orrick Apr 20 '14 at 15:39