Consider the following frequency distribution the lower limit of the median class is

Text Solution

1717.51818.5

Answer : B

Solution : <img src="https://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:

The upper limit of the median class is ______.

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.

`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. 

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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