Question

The class marks of a distribution area:-
37, 42, 47, 52, 57, 62, 67, 72
Determine the (i) class size (ii) class limits (iii) true class limits.

Answer 1

Hint: Take the difference between any two consecutive terms given in class marks to determine the class size. To find the class limits, apply the formula – lower class limit = class mark - $ \dfrac{\text{class size}}{2} $ and upper class limit = class mark + $ \dfrac{\text{class size}}{2} $ . Using these two formulas find the class limits and upper-class limits as intervals to determine the true class limits. Form a distribution table.

Complete step by step answer:
Here, we have been provided with certain class marks of distribution and we have to determine class size, class limits and true class limits. First, let us see the meaning of these terms.
Now, class size is there any friends between any two consecutive class marks of a distribution. So, considering the distribution provided in the question, we have,
Class marks:- 37, 42, 47, 52, 57, 62, 67, 72.
Here, let us select 37 and 42 as the two consecutive class marks. Therefore, we have,
(i) Class size = 42 – 37 = 5
Hence, the class size is 5.
(ii) Now, when we talk about class limits, they are of two types namely lower class limit and upper class limit. Lower class limit denotes the minimum value of data that can be included in an interval while upper class limit denotes the maximum value of data that can be entered in that interval. To find these 2 types of class limits, we use the formulas:-
Lower class limit = class mark - $ \dfrac{\text{class size}}{2} $
Upper class limit = class mark + $ \dfrac{\text{class size}}{2} $
So, let us calculate lower and upper class limits (L.C.L and U.C.L) for each class mark. Therefore, we have
For class mark = 37,
  $ \begin{align}
  & L.C.L=37-\dfrac{5}{2}=34.5 \\
 & U.C.L=37+\dfrac{5}{2}=39.5 \\
\end{align} $
For class mark = 42,
  $ \begin{align}
  & L.C.L=42-\dfrac{5}{2}=39.5 \\
 & U.C.L=42+\dfrac{5}{2}=44.5 \\
\end{align} $
For class mark = 47,
  $ \begin{align}
  & L.C.L=47-\dfrac{5}{2}=44.5 \\
 & U.C.L=47+\dfrac{5}{2}=49.5 \\
\end{align} $
For class mark = 52,
  $ \begin{align}
  & L.C.L=52-\dfrac{5}{2}=49.5 \\
 & U.C.L=52+\dfrac{5}{2}=54.5 \\
\end{align} $
For class mark = 57,
  $ \begin{align}
  & L.C.L=57-\dfrac{5}{2}=54.5 \\
 & U.C.L=57+\dfrac{5}{2}=59.5 \\
\end{align} $
For class mark = 62,
  $ \begin{align}
  & L.C.L=62-\dfrac{5}{2}=59.5 \\
 & U.C.L=62+\dfrac{5}{2}=64.5 \\
\end{align} $
For class mark = 67,
  $ \begin{align}
  & L.C.L=67-\dfrac{5}{2}=64.5 \\
 & U.C.L=67+\dfrac{5}{2}=69.5 \\
\end{align} $
For class mark = 72,
  $ \begin{align}
  & L.C.L=72-\dfrac{5}{2}=69.5 \\
 & U.C.L=72+\dfrac{5}{2}=74.5 \\
\end{align} $
(iii) Now we have to determine the true class limits. True class limits are obtained when these lower and upper-class limits are written in intervals. So, let us form a distribution table for this,

Class Marks	Lower Class Limit (L.C.L)	Upper Class Limit (U.C.L)	True Class
37	34.5	39.5	34.5 – 39.5
42	39.5	44.5	39.5 – 44.5
47	44.5	49.5	44.5 – 49.5
52	49.5	54.5	49.5 – 54.5
57	54.5	59.5	54.5 – 59.5
62	59.5	64.5	59.5 – 64.5
67	64.5	69.5	64.5 – 69.5
72	69.5	74.5	69.5 – 74.5

Note:
 You must draw a table for the given distribution so that everything can be understood easily. Remember the definitions and formulas to determine the class limits and class marks. You may note that in the table, the upper-class limit of any class mark is equal to the lower class limit of the next class mark. This is proof that our answer is correct.