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Surface Areas And Volumes
EXERCISE 13.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Lifetimes (in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Frequency 10 35 52 61 38 29 Determine the modal lifetimes of the components.

Step-by-Step Solution

Key Concept: The modal class is the class interval having the highest frequency. For grouped data, the mode (modal value) is estimated using the formula: $$L = L_0 + \frac{f_1 - f_0}{(2f_1 - f_0 - f_2)}\,h$$ where \(L_0\) is the lower limit of the modal class, \(f_1\) is its frequency, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively, and \(h\) is the class width.
1. Identify the frequencies for each class:
- 0‑20 : 10
- 20‑40 : 35
- 40‑60 : 52
- 60‑80 : 61
- 80‑100 : 38
- 100‑120 : 29

2. Find the modal class – the class with the greatest frequency. The maximum frequency is 61, which belongs to the class 60‑80.

3. Write down the required quantities for the mode formula:
- Lower limit of modal class, \(L_0\) = 60
- Class width, \(h\) = 20 (since each class interval is of equal width)
- Frequency of modal class, \(f_1\) = 61
- Frequency of preceding class, \(f_0\) = 52 (class 40‑60)
- Frequency of succeeding class, \(f_2\) = 38 (class 80‑100)

4. Apply the mode formula for grouped data:
$$L = 60 + \frac{61 - 52}{2\times61 - 52 - 38}\times 20$$
$$L = 60 + \frac{9}{122 - 90}\times 20$$
$$L = 60 + \frac{9}{32}\times 20$$
$$L = 60 + 0.28125 \times 20$$
$$L = 60 + 5.625$$
$$L = 65.625\text{ hours}$$

5. State the modal lifetime (rounded to one decimal place):
$$\boxed{\text{Modal lifetime } \approx 65.6 \text{ hours}}$$

Correct Answer: Modal lifetime ≈ 65.6 hours
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