Question

फाइंड मीन मेडियन एंड मॉड 100 200 323 को ट्वेंटी-20 8333​

Answer 1

Answer:

Class interval Mid value

x

i

Frequency

f

i

f

i

x

i

cf

0−50 35 2 50 2

50−100 75 3 225 5

100−150 125 5 625 10

150−200 175 6 1050 16

200−250 225 5 1127 21

250−300 275 3 825 24

300−350 325 1 325 25

∑f

i

=25 ∑f

i

x

i

=4225

⇒ Mean=

∑f

i

∑f

i

x

i

=

25

4225

=169

⇒ We have N=25$. Then,

2

N

=12.5

⇒ So, median class is 150−200.

l= lower limit of the modal class

h= size of the class intervals

f= frequency of the modal class

f

1

= frequency of the class preceding the modal class

f

2

= frequency of the class succeed in the modal class.

∴ l=150,h=200−150=50,f=6,cf=10

⇒ Median=l+

f

2

N

−cf

×h

⇒ Median=150+

6

12.5−10

×50

⇒ Median=150+

6

125

∴ Median=150+20.83=170.83.

⇒ Here maximum frequency is 6, then the corresponding class 150−200 is the modal class.

⇒ l=150,h=50,f=6,f

1

=5f

2

=5

⇒ Mode=l+

2f−f

1

−f

2

f−f

1

×h

⇒ Mode=150+

2×6−5−5

6−5

×50

⇒ Mode=150+

2

50

∴ Mode=150+25=175