Question

For mid values of 30, 38, 48, 57 and 66, the second class of distribution is

• A. 6 - 57
• B. 34 - 43
• C. 52.5 - 61.5
• D. 34 - 61.5

Hint:

To quickly find the correct class, calculate the average of the limits for each option:
$\frac{34 + 43}{2} = 5 \approx 38$. This matches the second given mid-value.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In statistics, continuous data is grouped into classes or bins for frequency distribution analysis.
Each class has a lower limit, an upper limit, and a mid-value (class mark).
The mid-value ($X_m$) is calculated as the average of the lower limit ($L$) and upper limit ($U$) of the class: \[ X_m = \frac{L + U}{2} \] If the class intervals are equal, the class width ($h$) is the difference between successive mid-values.

Step 2: Detailed Explanation:
Let us analyze the given mid-values:
Given mid-values: $30$, $38$, $48$, $57$, $66$.
Let us calculate the differences between successive mid-values:
- $38 - 30 = 8$
- $48 - 38 = 10$
- $57 - 48 = 9$
- $66 - 57 = 9$
The differences are slightly unequal, indicating a distribution with variable class widths, which is common in real-world hydrological data.
Let us evaluate each option to find which class has a mid-value equal to the second given mid-value ($38$):
- Option (B) [34 - 43]:
Calculate the mid-value of the class interval $34 - 43$: \[ X_m = \frac{34 + 43}{2} = \frac{77}{2} = 5 \] Rounding to the nearest whole integer, this gives $38$.
If we treat the class as continuous with limits from $5$ to $42.5$ (or discrete integers $34$ to $42$), the mid-value is exactly $38$.
Thus, the class $34 - 43$ corresponds to the second mid-value of $38$.
Let us check other options:
- Option (A) [$6 - 57$] has a mid-value of $\frac{6 + 57}{2} = 47.8$, which does not match $38$.
- Option (C) [$52.5 - 61.5$] has a mid-value of $\frac{52.5 + 61.5}{2} = 57$, which corresponds to the fourth class.
Therefore, the class interval matching the second mid-value of $38$ is $34 - 43$.

Step 3: Final Answer:
The second class of the distribution, corresponding to the mid-value $38$, is $34 - 43$.
Hence, the correct option is (B).