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Surface Areas And Volumes
EXERCISE 13.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure : Expenditure (in `) Number of families 1000 - 1500 24 1500 - 2000 40 2000 - 2500 33 2500 - 3000 28 3000 - 3500 30 3500 - 4000 22 4000 - 4500 16 4500 - 5000 7 STATISTICS 187

Step-by-Step Solution

Key Concept: For grouped data, the <b>modal class</b> is the class with the highest frequency. The mode is estimated using the formula \(\displaystyle \text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h\), where \(L\) is the lower limit of the modal class, \(f_1\) its frequency, \(f_0\) the frequency of the preceding class, \(f_2\) the frequency of the succeeding class and \(h\) the class width. The <b>mean</b> for grouped data is obtained by taking the class mid‑points as representative values, multiplying each by its frequency, summing these products and dividing by the total number of observations.
1. Identify the modal class\
The frequencies are: 24, 40, 33, 28, 30, 22, 16, 7.\
The largest frequency is 40, so the modal class is \(1500\!\!\text{–}\!2000\).

2. Compute the mode (grouped data formula)\
\[\begin{aligned}
L &= 1500 \text{ (lower limit of modal class)}\\
f_1 &= 40 \text{ (frequency of modal class)}\\
f_0 &= 24 \text{ (frequency of previous class)}\\
f_2 &= 33 \text{ (frequency of next class)}\\
h &= 2000-1500 = 500 \text{ (class width)}\\[4pt]
\text{Mode} &= L + \frac{f_1-f_0}{2f_1-f_0-f_2}\,h \\
&= 1500 + \frac{40-24}{2\times40-24-33}\times500 \\
&= 1500 + \frac{16}{80-57}\times500 \\
&= 1500 + \frac{16}{23}\times500 \\
&= 1500 + 0.695652\times500 \\
&= 1500 + 347.83 \approx \mathbf{1848\text{ (Rs.)}}
\end{aligned}\]
Hence the modal monthly expenditure is approximately Rs. 1850.

3. Find the mean expenditure\
- Compute the class mid‑points (\(x_i\)):\
\[\begin{array}{c|c}
\text{Class} & \text{Mid‑point } x_i \\ \hline
1000-1500 & 1250 \\
1500-2000 & 1750 \\
2000-2500 & 2250 \\
2500-3000 & 2750 \\
3000-3500 & 3250 \\
3500-4000 & 3750 \\
4000-4500 & 4250 \\
4500-5000 & 4750 \\
\end{array}\]
- Multiply each mid‑point by its frequency (\(f_i\)) and sum:
\[\begin{aligned}
\sum f_i x_i &= 1250\times24 + 1750\times40 + 2250\times33 + 2750\times28 \\
&\quad + 3250\times30 + 3750\times22 + 4250\times16 + 4750\times7 \\
&= 30\,000 + 70\,000 + 74\,250 + 77\,000 \\
&\quad + 97\,500 + 82\,500 + 68\,000 + 33\,250 \\
&= 532\,500
\end{aligned}\]
- Total number of families \(N = 200\).
- Mean \(\bar{x}\):
\[\bar{x}=\frac{\sum f_i x_i}{N}=\frac{532\,500}{200}=\mathbf{2\,662.5\text{ (Rs.)}}\]
- Rounded to the nearest rupee, the mean monthly expenditure is Rs. 2663.

4. Answer\
- Modal monthly expenditure ≈ Rs. 1850.\
- Mean monthly expenditure ≈ Rs. 2663.

Correct Answer: Modal expenditure ≈ Rs. 1850; Mean expenditure ≈ Rs. 2663.
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