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?
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