Find the missing frequencies and the median for the following distribution if the mean is 1.46.


No. of accidents:



0



1



2



3



4



5



Total



Frequency (No. of days):



46



?



?



25



10



5



200



No. of accidents (x)



No. of days (f)



Fx



0



46



0



1



X



X



2



Y



2y



3



25



75



4



10



40



5



5



25




N = 200



fx = x + 2y + 140



Given, N = 200


= 46 + x + y + 25 + 10 + 5 = 200


= x + y = 200 – 46 – 25 – 10 – 5


= x + y = 114 (i)


And Mean = 1.46


= 1.46


= = 1.46


= x + 2y + 140 = 292


= x + 2y = 292 – 140


= x + 2y = 152 (ii)


Subtract (i) from (ii), we get


X + 2y – x – y = 152 – 114


y = 38


Put the value of y in (i), we get


x = 114 – 38 = 76


No. of accidents



No. of days



Cumulative frequency



0



46



46



1



76



122



2



38



160



3



25



185



4



10



195



5



5



200




N = 200




We have, N = 200


= = 100


The cumulative frequency just more than is 122 so the median is 1


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