Answer :
Given Median =46
Then, median Class = 40 – 50
the lower limit (l) = 40
cumulative frequency of the class preceding 40 – 50 (cf) = 42 + x
frequency of the median class 40 – 50 = 65,
class size (h) = 10
Total frequencies (n) = 229
So, 150 + x + y = 229
⇒ x + y = 229 – 150
⇒ x + y = 79 …(i)
Using the formula,,we have
⇒39 = 72.5 – x
⇒ x = 33.5
Putting the value of x in eq. (i), we get
⇒ 33.5 + y = 79
⇒ y = 79 – 33.5
⇒ y = 45.5
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation


RELATED QUESTIONS :
A data has 25 obs
RS Aggarwal - Mathematics