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Topic Question Set

Q 1 :    

Let M denote the median of the following frequency distribution

Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20
Frequency     3     9     10     8     6

 

Then 20 M is equal to                                                                                                         [2024]

  • 416

     

  • 52

     

  • 208

     

  • 104

     

(3)

Class Frequency Cumulative Frequency
0 - 4             3                                    3
4 - 8             9                                   12
8 - 12             10                                    22
12 - 16              8                                    30
16 - 20              6                                    36
               36  

 

Where, h=4, l=8, c.f.=12, f=10, n=18

M=l+[n2-c.f.f]×h=8+18-1210×4

=8+6×410=10.4

Now, 20M=20×10.4=208

Q 2 :    

Marks obtains by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is          [2025]

  • 48

     

  • 52

     

  • 40

     

  • 44

     

(4)

In the given question, l = 12, cf = 18, median = 14, f = 12 and h = 6

∴  Median = l+(N2–cff)×h

⇒ 14=12+(N2–1812)×6

⇒ 2=N2–182 ⇒ 4=N2–18

⇒ N2=22 ⇒ N=44

So, total number of students = 44.