NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 13: Statistics and Probability - Exercise 13.3Question 11
Class 10 - Mathematics - CHAPTER 13: Statistics and Probability - Exercise 13.3
Question. 11
11. From the following “more than or equal to” data, form the frequency distribution:
| Marks (out of 90) | ≥80 | ≥70 | ≥60 | ≥50 | ≥40 | ≥30 | ≥20 | ≥10 | ≥0 |
|---|---|---|---|---|---|---|---|---|---|
| No. of candidates | 4 | 6 | 11 | 17 | 23 | 27 | 30 | 32 | 34 |
Answer:
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | 70–80 | 80–90 |
|---|---|---|---|---|---|---|---|---|---|
| Frequency | 2 | 2 | 3 | 4 | 6 | 6 | 5 | 2 | 4 |
Detailed Answer with Explanation:
Step 1: Look at the given table. It is in “more than or equal to” form. This means: for each mark limit, the number tells us how many students scored at least those marks.
Step 2: To convert into a normal frequency table (0–10, 10–20, etc.), we must subtract successive values.
Step 3: Calculate frequencies one by one.
- For 0–10: Total candidates (≥0) – candidates (≥10) = 34 – 32 = 2
- For 10–20: Candidates (≥10) – candidates (≥20) = 32 – 30 = 2
- For 20–30: Candidates (≥20) – candidates (≥30) = 30 – 27 = 3
- For 30–40: Candidates (≥30) – candidates (≥40) = 27 – 23 = 4
- For 40–50: Candidates (≥40) – candidates (≥50) = 23 – 17 = 6
- For 50–60: Candidates (≥50) – candidates (≥60) = 17 – 11 = 6
- For 60–70: Candidates (≥60) – candidates (≥70) = 11 – 6 = 5
- For 70–80: Candidates (≥70) – candidates (≥80) = 6 – 4 = 2
- For 80–90: Candidates (≥80) – candidates (≥90). Since ≥90 is not given, we assume it is 0. So, 4 – 0 = 4
Step 4: Write all these frequencies in the table beside their class intervals.
Final Note: Always remember: when data is given in “more than or equal to” form, you can find the actual frequency by subtracting successive entries.