# 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) $$\therefore\qquad\bar x = -2\times\frac12+1+2\times 3-\frac22=5$$

How do you define $u_i$? – dani_s Apr 19 '14 at 14:35