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+1 vote
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in Statistics by (30.8k points)
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Find the mean, median and mode of the following data:

2 Answers

+1 vote
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Classes xi​  fi​  fi​xi​  c.f. 
 0−20 10 60   6
 20−40 30 240   14
 40−60 50 10  500  24
 60−80 70 12 840  36
 80−100 90  540   42
 100−120 110 550   47
 120−140 130  390   50
 ∑fi ​= 50 ∑fi​xi ​= 3120 

⇒  Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac {3120}{50} = 62.4\)

⇒  N = 50 and \(\frac N2 = 25\)

∴  Median class = 60 − 80

l = lower limit of the modal class

h = size of the class intervals

f = frequency of the modal class

f1​ = frequency of the class preceding the modal class

f2 ​= frequency of the class succeed in the modal class.

⇒  Here, l = 60, f = 12, cf = 24, h = 20

Median = \(l + \frac{\frac N2- cf}f \times h\)

\(= 60 + \frac{25 - 24}{12} \times 20\)

\(= 60 + \frac{20}{12} \)

\(= 61.66\)

Modal class = 60 − 80

⇒  l = 30, f = 12, f1​ = 10, f2​ = 6, h = 20

Mode = \(l + \frac{f - f_1}{2f - f_1 - f_2}\times h\)

\(= 60 + \frac{12 - 10}{2\times 12 - 10 - 6}\times 20\)

\(= 60+ \frac 2{8} \times 20\)

\(= 60 + 5\)

\(= 65\)

+2 votes
by (31.2k points)

Mean = \(\frac{\sum fx}{N}=\frac{320}{50}=62.4\)

We have, 

N= 50,

N/2 = 25

Hence, 

median class =60-80, such that

l=60, f’=12, f=24, h=20

Median \(=I+\frac{\frac{n}{2}-f}{f^{1}}\times h\)

\(=60+\frac{25-24}{12}\times20\)

\(=60+1.67=61.67\)

Here, 

we may observe that maximum class frequency is 12 belonging to the class interval 60-80 

So, 

modal class= 60-80 

Lower limit, l= 60 

f0=10, f2=6, f=12,h = 20

Mode \(=I+(\frac{f-f_0}{2f-f_0-f_2})h\)

\(=60+(\frac{12-10}{24-10-6})20\)

\(=60+\frac{40}{8}=65\)

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