Text Solution
1717.51818.5
Answer : B
Solution : <img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_X_C13_S01_007_S01.png" width="80%"> <br> Given, classes are not continous, so we make continuous by substracting 0.5 from lower limit and adding 0.5 to upper limit of each class <br> Here, `(N)/(2)=(57)/(2)=28.5` which lies in the interval 11.5-17-5. Hence the upper limit is 17.5
Consider the following frequency distribution:
Class | 0 – 5 | 6 – 11 | 12 – 17 | 18 – 23 | 24 – 29 |
Frequency | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is ______.
Class | 0 – 5 | 6 – 11 | 12 – 17 | 18 – 23 | 24 – 29 |
Frequency | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is 17.5.
Explanation:
Classes are not continuous
Hence, we make the data continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.
Class | Frequency | Cumulative Frequency |
0.5 – 5.5 | 13 | 13 |
6.5 – 11.5 | 10 | 23 |
11.5 – 17.5 | 15 | 38 |
17.5 – 23.5 | 8 | 46 |
23.5 – 29.5 | 11 | 57 |
`N/2 = 57/2 = 28.5`
28.5 lies in between the interval 11.5 – 17.5.
Therefore, the upper limit is 17.5
Concept: Median of Grouped Data
Is there an error in this question or solution?
The given classes in the table are non-continuous. So, we first make the classes continuous by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit in each class.
Class | Frequency | Cumulative Frequency |
0.5–5.5 | 13 | 13 |
5.5–11.5 | 10 | 23 |
11.5–17.5 | 15 | 38 |
17.5–23.5 | 8 | 46 |
23.5–29.5 | 11 | 57 |
Now, from the table we see that N = 57.
So,
\[\frac{N}{2} = \frac{57}{2} = 28 . 5\]
28.5 lies in the class 11.5–17.5.The upper limit of the interval 11.5–17.5 is 17.5.
Hence, the correct answer is option (b).
Consider the following frequency distribution :[ Class: 0 5 6 11 12 17 18 23; Frequency: 13 10 15 8 24 29; ]The upper limit of the median class isa 17b 17.5c 18d 18.5
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