# Given arithmetic mean and standard deviation for a frequency distribution, the actual class intervals are required

The Text I am reading has a question that goes like this :

Explain clearly the ideas implied in using arbitrary working origin and scale for the calculation of arithmetic mean and standard deviation of a frequency distribution. The values of the arithmetic mean and standard deviation of the following frequency distribution of a continuous variable derived from the analysis in the above manner are $40.604\,\text{lb}$ and $7.92\,\text{lb}$. Determine the actual class intervals.

$$\begin{array}{crrrrrrrr} x: & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ f: & 3 & 15 & 45 & 57 & 50 & 36 & 25 & 9 \end{array}$$

Please correct me if I am wrong, I think that the change in scale and origin would be required if some of the meaningless values have to be removed from the distribution, like in the distribution above the unit we are considering is $lb$ which is related to weight and hence values $-3,-2,-1,0$ are problematic. Hence, the transformation $y_i = 5-x_i$ would make all the values greater than equal to one, but still what about the class intervals ?

Can anyone help ?

• Please also add the self-study tag and read its tag wiki – Glen_b -Reinstate Monica May 16 '17 at 23:35
• @User9523 I think there was a typo in the x-values. – L.V.Rao May 17 '17 at 4:01

Let $y_{i}$ denote the mid-values of the original grouped frequency table, $A$ be the arbitrarily chosen origin, and $C$ denote the class-width. Considering the transformation $\dfrac{y_{i}-A}{C}$ for such a table, we obtained $$\begin{array}{c|cccccccc} x_{i}&-3&-2&-1&0&1&2&3&4\\ \hline f_{i}&3&15&45&57&50&36&25&9 \end{array}$$
\begin{eqnarray*} \bar{X}&=&\dfrac{\sum f_{i}x_{i}}{\sum f_{i}}=\dfrac{1}{\sum f_{i}}\sum f_{i}\left(\dfrac{y_{}-A}{C}\right)=\dfrac{1}{C}\left( \dfrac{\sum f_{i}y_{i}}{\sum f_{i}}-A \right)\\ \bar{Y}&=& C\bar{X} + A \qquad\cdots\qquad (i) \end{eqnarray*} Equation $(i)$ involves two unknowns $C$ and $A$.
Variance of original data $y_{i}-A$= $d_{i}$ is defined by \begin{eqnarray*} \dfrac{\sum f_{i}d_{i}^{2}}{\sum f_{i}}-\left(\dfrac{\sum f_{i}d_{i}}{\sum f_{i}} \right)^2 &=& \dfrac{C^2\sum f_{i}x_{i}^{2}}{\sum f_{i}}-\left(\dfrac{C\sum f_{i}x_{i}}{\sum f_{i}} \right)^2\\ SD_{y} &=& C \sqrt{ \dfrac{\sum f_{i}x_{i}^{2}}{\sum f_{i}}-\left(\dfrac{\sum f_{i}x_{i}}{\sum f_{i}} \right)^2}\qquad\cdots\qquad (ii) \end{eqnarray*} where the quantity under the radical sign is the $SD$ of the transformed data, given in the table above. As the standard deviation $SD_y$ of the data is given, equation $(ii)$ involves only one unknown, that is $C$. So, from the given data table just compute the $SD$ and plug in the known $SD_y$ value son the LHS to obtain $C$. Use this $C$ value in above equation $(i)$ for mean to obtain the other unknown $A$.