Question

The following table gives the distribution of 100 families according to daily expenditure:
Expenditure (in Rs): 0-10, 10-20, 20-30, 30-40, 40-50
Number of families: 14, x, 27, y, 15
If the mode of the distribution is 24, find the missing frequencies x and y.

• A. x = 16, y = 20
• B. x = 18, y = 21
• C. x = 23, y = 21
• D. x = 21, y = 23

Hint:

For mode in grouped data, identify the modal class first.
Use the mode formula carefully, and also use the total frequency to find missing frequencies.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The mode is the value that occurs most frequently.
In a grouped frequency distribution, the modal class is the class with the highest frequency.

Step 2: Key Formula or Approach:
For mode in a grouped frequency distribution: \[ \text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h, \]
where \(L\) is the lower limit of the modal class, \(h\) is the class width,
\(f_1\) is the frequency of the modal class, \(f_0\) is the frequency of the class before, and \(f_2\) is the frequency of the class after.

Step 3: Detailed Explanation:
Given mode = 24, which lies in the class 20-30.
So modal class is 20-30.
\(L = 20\), \(h = 10\), \(f_1 = 27\), \(f_0 = x\), \(f_2 = y\).
Using the formula: \[ 24 = 20 + \frac{27 - x}{2(27) - x - y} \times 10. \]
Simplify: \[ 4 = \frac{27 - x}{54 - x - y} \times 10 \implies 0.4 = \frac{27 - x}{54 - x - y}. \]
Cross-multiply: \[ 0.4(54 - x - y) = 27 - x. \]
\[ 21.6 - 0.4x - 0.4y = 27 - x. \]
\[ x - 0.4x - 0.4y = 27 - 21.6. \]
\[ 0.6x - 0.4y = 5.4. \]
Multiply by 10: \[ 6x - 4y = 54 \implies 3x - 2y = 27. \quad (1) \]
Also total families = 100: \[ 14 + x + 27 + y + 15 = 100 \implies x + y = 44. \quad (2) \]
Solve equations (1) and (2): From (2), \(x = 44 - y\).
Substitute in (1): \[ 3(44 - y) - 2y = 27 \implies 132 - 3y - 2y = 27 \implies 132 - 5y = 27 \implies -5y = -105 \implies y = 21. \]
Then \(x = 44 - 21 = 23\).
Thus, \(x = 23\), \(y = 21\), which is option (C).