Question

The median for the following frequency distribution is 34 and the sum of all frequencies is 80.

Class Interval

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

Frequency

3

x

16

30

y

5


What are the values of x and y?

 

8, 16

 

8, 18

 

9, 16

 

9, 17


​

Answer 1

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Answer 2

Answer:

The missing values x = 9 and y = 17

Step-by-step explanation:

Given data

Class Interval   0 – 10   10 – 20    20 – 30   30 – 40   40 – 50   50 – 60  

Frequency            3            x              16             30              y            5

⇒ The median of the following data = 34

⇒ sum of the frequencies = 80

here we need to find missing values of x and y

⇒ To find x and y we will calculate median of the given data for that we need to calculate  cumulative frequencies for given data

cumulative frequency table  

  class interval             frequency                 cumulative frequencies

      0 – 10                          3                                3            

     10 – 20                         x                                3+ x

     20 – 30                        16                              19+x

     30 – 40                        30                              49+x

     40 – 50                         y                               49+x+y

     50 – 60                        5                               54+x+y  

⇒ from given data sum frequencies = 80

⇒  54 + x+ y = 80

⇒ x + y = 26 _ (1)

⇒ median of the data is 34 which is lies in 30 - 40  

⇒ therefore in given data 30 - 40 is the median class  

the formula for median is given by median =  $l + [ \frac{ \frac{N}{2} - CF }{f} ]h$  

here

 l = lower limit of median class  = 30

 N = sum of frequencies = 80  [from given data]

 C.F = cumulative frequency of preceding class of median class = 19+x

 f = median class frequency  = 30

 h = size of class = 10

⇒ median  = 30 + $[\frac{\frac{80}{2} - 19 - x }{30} ] 10$ = 34

                 ⇒       $[\frac{40-19-x}{30} ] 10 = 34 -30$      

                 ⇒     (21 - x)/ 3 = 4

                 ⇒      21-  x = 12

                 ⇒       - x  = 12 -21

                 ⇒        - x = - 9  

                 ⇒          x = 9  

⇒ substitute x = 9 in (1) ⇒ 9 + y = 26

                                                y = 26 - 9 = 17

⇒  the missing values x = 9 and y = 17