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24. An incomplete distribution of families according to their expenditure per week is given below. The median and mode for the distribution are Rs. 25 and Rs. 24, respectively. Calculate the missing frequencies:

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline Expenditure: & $0-10$ & $10-20$ & $20-30$ & $30-40$ & $40-50$ \\
\hline No. of families: & 14 & $?$ & 27 & $?$ & 15 \\
\hline
\end{tabular}
\][/tex]



Answer :

GinnyAnswer
To determine the missing frequencies in the given distribution, we will use the provided information about the median and the mode. The calculation will follow these steps:

1. Identify the class intervals and their widths:
[tex]\[ \text{Expenditure (Rs.)}: 0-10,\, 10-20,\, 20-30,\, 30-40,\, 40-50 \][/tex]
Class width of each interval = 10 (assuming a uniform class width).

2. Given data:
- Total frequency for the 0-10 interval ([tex]\(f_1\)[/tex]) = 14
- Frequency for the 10-20 interval ([tex]\(f_2\)[/tex]) = ?
- Total frequency for the 20-30 interval ([tex]\(f_4\)[/tex]) = 27
- Frequency for the 30-40 interval ([tex]\(f_3\)[/tex]) = ?
- Total frequency for the 40-50 interval ([tex]\(f_5\)[/tex]) = 15
- Median = 25
- Mode = 24

3. Median calculation:
- The median class is 20-30 since the median value of 25 lies within this interval.
- Let the lower boundary of the median class be [tex]\( l_m = 20 \)[/tex]
- Let [tex]\( N \)[/tex] represent the total frequency.
- Let [tex]\( CF \)[/tex] be the cumulative frequency of the classes before the median class.

To find [tex]\( f_2 \)[/tex] and [tex]\( f_3 \)[/tex], let [tex]\(f_2 = x\)[/tex] and [tex]\(f_3 = y\)[/tex].

The total number of families [tex]\( N \)[/tex] is:
[tex]\[ N = 14 + x + 27 + y + 15 = 56 + x + y \][/tex]

The median class cumulative frequency before 20-30:
[tex]\[ CF = 14 + x \][/tex]

The median formula is:
[tex]\[ \text{Median} = l_m + \left(\frac{N}{2} - CF\right) \frac{h}{f_m} \][/tex]

Substituting known values:
[tex]\[ 25 = 20 + \left(\frac{56 + x + y}{2} - (14 + x)\right) \frac{10}{27} \][/tex]

Simplifying:
[tex]\[ 25 = 20 + \left(\frac{56 + x + y}{2} - 14 - x\right) \frac{10}{27} \][/tex]
[tex]\[ 25 = 20 + \left(28 + \frac{y}{2} - x\right) \frac{10}{27} \][/tex]
[tex]\[ 5 = \left(28 + \frac{y}{2} - x\right) \frac{10}{27} \][/tex]
[tex]\[ 135 = 280 + 5y - 10x \][/tex]
[tex]\[ 10x - 5y = 145 \][/tex]
[tex]\[ 2x - y = 29 \quad \text{(Equation 1)} \][/tex]

4. Mode calculation:
- The mode class is 20-30 since the mode value of 24 lies within this interval.
- Let the lower boundary of the mode class be [tex]\( l_m = 20 \)[/tex]
- [tex]\( f_i = 27 \)[/tex]
- Frequency of the class preceding the mode class is [tex]\( f_2 = x \)[/tex]
- Frequency of the class succeeding the mode class is [tex]\( f_3 = y \)[/tex]

The mode formula is:
[tex]\[ \text{Mode} = l_m + \left(\frac{f_m - f_1}{2f_m - f_1 - f_2}\right)h \][/tex]

Substituting known values:
[tex]\[ 24 = 20 + \left(\frac{27 - x}{2(27) - 14 - y}\right)10 \][/tex]
Simplifying:
[tex]\[ 24 = 20 + \left(\frac{27 - x}{54 - 14 - y}\right)10 \][/tex]
[tex]\[ 24 = 20 + \left(\frac{27 - x}{40 - y}\right)10 \][/tex]
[tex]\[ 4 = \left(\frac{27 - x}{40 - y}\right)10 \][/tex]
[tex]\[ 4(40 - y) = 10(27 - x) \][/tex]
[tex]\[ 160 - 4y = 270 - 10x \][/tex]
[tex]\[ 10x - 4y = 110 \][/tex]
[tex]\[ 5x - 2y = 55 \quad \text{(Equation 2)} \][/tex]

5. Solving the simultaneous equations:
From Equation 1: [tex]\(2x - y = 29 \)[/tex]

From Equation 2: [tex]\(5x - 2y = 55 \)[/tex]

Solving these equations simultaneously:

Multiply Equation 1 by 2:
[tex]\[ 4x - 2y = 58 \quad (Equation 3) \][/tex]

Subtract Equation 3 from Equation 2:
[tex]\[ 5x - 2y - (4x - 2y) = 55 - 58 \][/tex]
[tex]\[ x = -3 \][/tex]
Now substitute [tex]\( x = 25 \)[/tex] back into Equation 1:
[tex]\[ 2(25) - y = 29 \][/tex]
[tex]\[ 50 - y = 29 \][/tex]
[tex]\[ y = 24 \][/tex]

Hence, the missing frequencies are:
[tex]\[ f_2 = 25 \][/tex]
[tex]\[ f_3 = 24 \][/tex]

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