VERIFICATION: Apply Step-deviation method to find the average mean of the following frequency distribution Q: Apply Step-deviation method to find the average mean of the following frequency distribution
\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|} \hline
Variate (x_i)  & 5  & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 \\ \hline
Frequency(f_i) & 20 & 43 & 75 & 67 & 72 & 45 & 39 & 9  & 8  & 6 \\ \hline
\end{array}

My Solution 
\begin{array}{|l|l|l|l|l|l|}
\hline
\textbf{h=10} & \textbf{$f_i$}  & \textbf{$xi$} & \textbf{$di=xi-a$ ; a=30} & \textbf{$ui=\frac{di}{a}$} & \textbf{$f_iu_i$}   \\
5-15                   & 63           & 10          & -20                     & -2               & -126            \\ \hline
15-25                  & 142          & 20          & -10                     & -1               & -142            \\ \hline
25-35                  & 117          & 30          & 0                       & 0                & 0               \\ \hline
35-45                  & 48           & 40          & 10                      & 1                & 48              \\ \hline
45-55                  & 14           & 50          & 20                      & 2                & 28              \\ \hline
                       & \Sigma f_i=384 &             &                         &                  & \Sigma f_iu_i=-192 \\ \hline
\end{array}
 So, $$\bar x = a + \left(\frac{\Sigma f_iu_i}{\Sigma f_i}\right)(h)$$
$$ \bar x= 30 + \left(\frac{-192}{384}\right)(10)  $$
$$\bar x = 30 -5$$
$$ \therefore \bar x = 25$$

but the correct answer turns out to be 22.214.

Where am i worng?
 A: The source of error is the wrong combination of the $2i-1$ and $2i$-th data into one class of wrong width $h=10$.  You're in fact calculating
$$\sum_{i=1}^5 x_{2i}(f_{2i-1}+f_{2i}) \tag{overestimated}$$
instead of
$$\bar x = \sum_{i=1}^5 (x_{2i-1}f_{2i-1} + x_{2i}f_{2i}) = \sum_{i=1}^{10} x_i f_i.$$
Since the common difference of successive terms in $x_i$ is $5$, we should choose $h=5$.
\begin{array}{|r|r|r|r|r|} \hline
\text{Marks } x_i & \text{freq. } f_i & {\text{Deviation } \\ d_i=x_i-a \\ a = 30} & {u_i=\frac{d_i}{h} \\ \text{Here } h = 5} & f_iu_i \\ \hline
  5 & 20 & -25 & -5 & -100 \\ \hline
 10 & 43 & -20 & -4 & -172 \\ \hline
 15 & 75 & -15 & -3 & -225 \\ \hline
 20 & 67 & -10 & -2 & -134 \\ \hline
 25 & 72 &  -5 & -1 &  -72 \\ \hline
 30 & 45 &   0 &  0 &    0 \\ \hline
 35 & 39 &   5 &  1 &   39 \\ \hline
 40 &  9 &  10 &  2 &   18 \\ \hline
 45 &  8 &  15 &  3 &   24 \\ \hline
 50 &  6 &  20 &  4 &   24 \\ \hline
 & \sum f_i=384 & & & \sum f_i u_i=-598 \\ \hline
\end{array}
\begin{align}
\bar x &= a + \left(\frac{\sum f_iu_i}{\sum f_i}\right)(h) \\
&= 30 + \left(\frac{-598}{384}\right)(5) \\
&= 30 - \frac{1495}{192} \\
&= \frac{4265}{192} \\
&= 22.2135 \tag{cor. to 4 d.p.}
\end{align}
Reference: https://gradestack.com/CBSE-Class-10th-Course/Statistic/Step-deviation-Method/15052-3000-5351-study-wtw
