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Surface Areas And Volumes
EXERCISE 13.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure 100 - 150 150 - 200 200 - 250 250 - 300 300 - 350 (in `) Number of 4 5 12 2 2 households Find the mean daily expenditure on food by a suitable method.

Step-by-Step Solution

Key Concept: For grouped data, the mean is obtained by taking the class‑midpoint (or assumed mean) of each interval, multiplying it by the frequency, summing these products and dividing by the total number of observations.
1. Identify the class intervals and their frequencies
\[\begin{array}{c|c}
\text{Class interval (₹)} & \text{Frequency (f)}\\ \hline
100-150 & 4\\
150-200 & 5\\
200-250 & 12\\
250-300 & 2\\
300-350 & 2\\
\end{array}\]
Total frequency \(N = 4+5+12+2+2 = 25\).

2. Find the class‑midpoint (x) for each interval
\[x = \frac{\text{lower limit}+\text{upper limit}}{2}\]
\[\begin{array}{c|c}
\text{Class interval} & \text{Midpoint (x)}\\ \hline
100-150 & 125\\
150-200 & 175\\
200-250 & 225\\
250-300 & 275\\
300-350 & 325\\
\end{array}\]

3. Compute \(f\times x\) for each class
\[\begin{array}{c|c|c}
\text{Midpoint (x)} & \text{Frequency (f)} & f\times x\\ \hline
125 & 4 & 500\\
175 & 5 & 875\\
225 & 12 & 2700\\
275 & 2 & 550\\
325 & 2 & 650\\
\end{array}\]
Sum of \(f\times x\): \(\sum f x = 500+875+2700+550+650 = 5275\).

4. Calculate the mean
\[\bar{x} = \frac{\sum f x}{N} = \frac{5275}{25} = 211\]

5. Interpretation
The average (mean) daily expenditure on food per household is \(\mathbf{₹\,211}\).

Correct Answer: ₹211
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