Q5:

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. 

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Given: The frequency distribution of mangoes in packing boxes is as follows:

To Find: The mean number of mangoes kept in a packing box. We will use the Step-Deviation Method as it simplifies calculations when class intervals are continuous and frequencies are large.

Step 1: Prepare the Frequency Distribution Table

First, we observe that the class intervals are not continuous (e.g., 52 to 53). We adjust them by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. Then, we calculate the class mark ($x_i$) for each interval using the formula: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

Step 2: Identify Variables for Step-Deviation Method

Let the assumed mean ($a$) = $57$.
The class size ($h$) = $52.5 - 49.5 = 3$.
Sum of frequencies ($\sum f_i$) = $400$.
Sum of product of frequencies and deviations ($\sum f_i u_i$) = $25$.

Step 3: Apply the Mean Formula

The formula for the mean ($\bar{x}$) using the Step-Deviation Method is:
$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$

Substituting the values:
$\bar{x} = 57 + \left( \frac{25}{400} \right) \times 3$

Step 4: Perform Arithmetic Calculations

$\bar{x} = 57 + \left( \frac{1}{16} \right) \times 3$ [Since $25/400 = 1/16$]
$\bar{x} = 57 + \frac{3}{16}$
$\bar{x} = 57 + 0.1875$
$\bar{x} = 57.1875$

Conclusion on Method: The Step-Deviation method was chosen because the class intervals are uniform and the values of $f_i$ and $x_i$ are large, making manual multiplication cumbersome. This method reduces the magnitude of the numbers, facilitating easier calculation.

Final Answer: The mean number of mangoes kept in a packing box is 57.19 (rounded to two decimal places).