plz guys help me to solve this question.
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Step-by-step explanation:
class interval. Frequency. C.F.
100-200. 11. 11
200-300. 12. 23
300-400. 10. 33
400-500. 13. 46
500-600. 20. 66
600-700. 14. 80
(n/2)th term
(80/2)th term
40th term
median= L+(n/2)-C.f.(p)/f ×i
400+(40)-33/80×100
400+8.75
408.75
So,the median is 408.75
Hope it will be help you
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