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In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000.Find the missing frequencies x and y.
Runs Scored     number of batsmen
2500-3500          5                     
3500-4500          x                     
4500-5500          y                     
5500-6500         12                    
6500-7500          6                     
7500-8500          2                     

 

Asked by ranjitrk1 15th June 2016, 10:31 PM
Answered by Expert
Answer:
begin mathsize 12px style Now comma space straight N space equals space sum for blank of straight f subscript straight i equals 60
rightwards double arrow 60 space equals space 25 plus straight x plus straight y
rightwards double arrow straight x plus straight y equals 35... left parenthesis straight i right parenthesis
Since space the space median space is space given space to space be space 5000 comma space it space lies space in space the space class space 4500 minus 5500.
rightwards double arrow straight l equals 4500 comma space straight h space equals space 1000 comma space straight f space equals space straight y comma space straight F space equals space 5 plus straight x space and space straight N space equals space 60
Median space equals space straight l space plus space fraction numerator begin display style straight N over 2 end style minus straight F over denominator straight f end fraction cross times straight h
rightwards double arrow 5000 space equals 4500 space plus space fraction numerator begin display style 60 over 2 minus left parenthesis 5 plus straight x right parenthesis end style over denominator straight y end fraction cross times 1000
rightwards double arrow 500 equals space fraction numerator begin display style 30 minus 5 minus straight x end style over denominator straight y end fraction cross times 1000
rightwards double arrow 500 straight y equals space left parenthesis 25 minus straight x right parenthesis 1000
rightwards double arrow straight y over 2 equals space left parenthesis 25 minus straight x right parenthesis
rightwards double arrow straight y equals 50 minus 2 straight x
rightwards double arrow 2 straight x space plus space straight y space equals space 50.... left parenthesis ii right parenthesis
from space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
2 straight x space plus space left parenthesis 35 space minus space straight x right parenthesis space equals space 50 rightwards double arrow straight x space equals space 15
Putting space straight x space equals space 15 space in space left parenthesis straight i right parenthesis comma space we space get space straight y space equals space 20.
Hence comma space straight x space equals space 15 space and space straight y space equals space 20. end style
Answered by Expert 17th June 2016, 10:34 AM
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