Question

❗❕❗100 pts question❗❕❗
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Answer 1

Answer:

BRO I AM SOLVING ONLY ONE YOU MIGHT SOLVE NEXT BY THAT METHOD OKKK.

Answer 2

Answer:

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Step-by-step explanation:

Here, the class intervals are of unequal width. If the class intervals are of unequal width the frequencies need not be adjusted to make the class intervals equal.

Calculation of Median

Marks No. of students (frequency) Cumulative frequency

0-10 5 5

10-30 15 20

30-60 30 50

60-80 8 58

80-90 2 60

N=∑f

i

=60

Here, N=60∴N/2=30

The cumulative frequency just greater than N/2=30 is 50 and the corresponding class is 30−60 . Hence, 30−60 is the median class.

∴l=30,f=30,F=20,h=30

Now, Median =l+

f

2

N

−F

×h

⇒Median=30+

30

30−20

×30=40