Question

The monthly expenditure on fruits in 200 families of a Housing Society is given below. Find the value of $x$ and also find the mode and mean expenditure on fruits.

Hint:

Using the step-deviation method for mean calculations with large class intervals saves you from dealing with large products ($f_i \cdot y_i$), keeping calculations rapid and error-free.

Answer 1

Step 1: Understanding the Question:
We are given a grouped frequency table of the monthly expenditure on fruits in 200 families.
One of the frequency values is unknown, labeled as $x$.
We need to find the value of $x$, and then calculate the mean and mode of the monthly expenditure.

Step 2: Key Formula or Approach:
1. Finding $x$: The sum of all frequencies must be equal to the total number of families (200).
2. Finding the Mean ($\bar{X}$): We will use the step-deviation method:
\[ \bar{X} = A + h \cdot \left( \frac{\sum f_i u_i}{\sum f_i} \right) \]
where $A$ is the assumed mean, $h$ is the class size ($500$), and $u_i = \frac{x_i - A}{h}$.
3. Finding the Mode:
\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
where $L$ is the lower limit of the modal class, $f_1$ is the frequency of the modal class, $f_0$ is the preceding frequency, and $f_2$ is the succeeding frequency.

Step 3: Detailed Explanation:

• 1. Calculate the value of $x$:
The total number of families is 200:
\[ 24 + 40 + 33 + 28 + x + 22 + 16 + 7 = 200 \]
\[ 170 + x = 200 \implies x = 30 \]

• 2. Calculate the Mean ($\bar{X}$):
Let us construct the step-deviation table with assumed mean $A = 2750$ and class width $h = 500$:
Now, apply the mean formula:
\[ \bar{X} = 2750 + 500 \cdot \left( \frac{-35}{200} \right) \]
\[ \bar{X} = 2750 - 2.5 \cdot 35 \]
\[ \bar{X} = 2750 - 87.5 = 2662.5\text{ Rupees} \]

• 3. Calculate the Mode:
- Find the modal class: The maximum frequency is 40, which lies in the class interval 1500-2000.
- Identify the variables for the mode formula:
- Lower limit, $L = 1500$
- Frequency of the modal class, $f_1 = 40$
- Frequency of the class preceding the modal class, $f_0 = 24$
- Frequency of the class succeeding the modal class, $f_2 = 33$
- Class width, $h = 500$
- Apply the formula:
\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
\[ \text{Mode} = 1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500 \]
\[ \text{Mode} = 1500 + \left( \frac{16}{80 - 57} \right) \times 500 \]
\[ \text{Mode} = 1500 + \frac{16}{23} \times 500 = 1500 + \frac{8000}{23} \]
\[ \text{Mode} \approx 1500 + 347.83 = 1847.83\text{ Rupees} \]


Step 4: Final Answer:
The value of $x$ is 30.
The mean expenditure on fruits is Rs. $2662.50$ and the mode is Rs. $1847.83$.