NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 13: Statistics and Probability - Exercise 13.3Question 6
Class 10 - Mathematics - CHAPTER 13: Statistics and Probability - Exercise 13.3
Question. 6
6. An aircraft has 120 seats. Over 100 flights, the seats occupied were:
| Seats | 100–104 | 104–108 | 108–112 | 112–116 | 116–120 |
|---|---|---|---|---|---|
| Frequency | 15 | 20 | 32 | 18 | 15 |
Determine the mean number of seats occupied.
Answer:
109.92 seats
Detailed Answer with Explanation:
Step 1: Identify the data.
We are given the number of seats occupied in groups (called classes) and how many times they occurred (called frequency).
| Class Interval (Seats) | 100–104 | 104–108 | 108–112 | 112–116 | 116–120 |
|---|---|---|---|---|---|
| Frequency (f) | 15 | 20 | 32 | 18 | 15 |
Step 2: Find the mid-point (class mark) of each interval.
The class mark is the average of the lower and upper limits:
- For 100–104: \( \tfrac{100+104}{2} = 102 \)
- For 104–108: \( \tfrac{104+108}{2} = 106 \)
- For 108–112: \( \tfrac{108+112}{2} = 110 \)
- For 112–116: \( \tfrac{112+116}{2} = 114 \)
- For 116–120: \( \tfrac{116+120}{2} = 118 \)
Step 3: Multiply frequency by class mark.
We calculate \( f \times x \):
| Class mark (x) | 102 | 106 | 110 | 114 | 118 |
|---|---|---|---|---|---|
| Frequency (f) | 15 | 20 | 32 | 18 | 15 |
| f × x | 1530 | 2120 | 3520 | 2052 | 1770 |
Step 4: Add the values.
- Total frequency: \( \sum f = 15+20+32+18+15 = 100 \)
- Total of \( f \times x \): \( \sum fx = 1530+2120+3520+2052+1770 = 10992 \)
Step 5: Apply the mean formula.
The mean is given by:
\[ \bar{x} = \dfrac{\sum f x}{\sum f} \]
Substituting the values:
\[ \bar{x} = \dfrac{10992}{100} = 109.92 \]
Final Answer: The mean number of seats occupied = 109.92 seats.