Class 0−44−88−1212−1616−20Frequency 391086Cumulative frequency 312223036 M=l+(f2N−C)h
Here, N=3+9+10+8+6=36
⇒2N=18
So, median class is (8−12)
⇒l=8,C=22,f=10,h=4
⇒M=8+1018−12×4
⇒M=10.4
⇒20M=208
Let M denote the median of the following frequency distribution.
Class Frequency 0−434−898−121012−16816−206
Then 20M is equal to :
Held on 30 Jan 2024 · Verified 6 Jul 2026.
416
104
52
208
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