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Surface Areas And Volumes
EXERCISE 13.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored Number of batsmen 3000 - 4000 4 4000 - 5000 18 5000 - 6000 9 6000 - 7000 7 7000 - 8000 6 8000 - 9000 3 9000 - 10000 1 10000 - 11000 1 Find the mode of the data.

Step-by-Step Solution

Key Concept: For grouped data, the mode is estimated using the formula: $$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$$ where \(L\) is the lower boundary of the modal class, \(f_1\) is the frequency of the modal class, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively, and \(h\) is the class width.
1. Identify the modal class – the class with the highest frequency.\
The frequencies are: 4, 18, 9, 7, 6, 3, 1, 1. Hence the modal class is 4000 – 5000.

2. Write down the required quantities:\
- Lower class boundary of the modal class, \(L = 4000\).
- Class width, \(h = 5000-4000 = 1000\).
- Frequency of modal class, \(f_1 = 18\).
- Frequency of the preceding class, \(f_0 = 4\) (class 3000–4000).
- Frequency of the succeeding class, \(f_2 = 9\) (class 5000–6000).

3. Apply the mode formula for grouped data:\
$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$$
Substituting the values:
$$\text{Mode}=4000+\frac{18-4}{2\times18-4-9}\times1000$$
$$\text{Mode}=4000+\frac{14}{36-13}\times1000$$
$$\text{Mode}=4000+\frac{14}{23}\times1000$$
$$\text{Mode}=4000+608.7$$
$$\text{Mode}\approx 4608.7$$

4. State the answer – Rounding to the nearest whole number (as runs are counted in whole numbers), the mode is approximately 4609 runs.

Thus, the most frequently occurring run‑score interval is centred around ≈ 4609 runs.

Correct Answer: ≈ 4609 runs (mode ≈ 4608.7)
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