Question

Find mean and mode of following
classes Frequency

0-20. 10

20-40 35

40-60 52

60-80. 61

80-100. 38

100-120 29
​

Answer 1

Answer:

Explanation:

SOLUTION :  

FREQUENCY DISTRIBUTION TABLE is in the attachment

For MODE :

Here the maximum frequency is 12, and the class corresponding to this frequency is 60 – 80. So the modal class is 60 - 80.

Therefore, l = 60, h = 20,  f1= 12,  f0= 10 , f2 = 6

Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h

= 60 + [(12 - 10)/(2 × 12 - 10 – 6) ] ×20

= 60 + [2 × 20)/(24 - 16)]

= 60 + [40/ 8]

= 60 + 5

= 65

MODE = 65

Hence, the mode of the data is 65 .

MEAN :  

From the table : Σfi = 50 , Σfixi = 3120

Mean = Σfixi /Σfi

Mean =  3120/50 = 312/5 = 62.4  

Hence, the Mean of the data is 62.4 .

For MEDIAN :  

Here, n = 50

n/2 = 25

Since, the Cumulative frequency just greater than 25  is 36 and the corresponding class is 60 - 80 .  Therefore 60 - 80 is the median class.

Here, l = 60  , f = 12 , c.f = 24,  h = 20

MEDIAN = l + [(n/2 - cf )/f ] ×h

= 60 + [(25 - 24)/12] × 20

= 60 + [(1 × 20)/12]

= 60 + 20/12

= 60 + 5/3  

= 60 + 1.66

= 61.66

Hence, the Median of the data is 61.66 .

★★ Mode = l + (f1-f0/2f1-f0-f2) ×h

l = lower limit of the modal class

h = size of the class intervals

f1 = frequency of the modal class

f0 = frequency of the class preceding the modal class

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

★★ MEDIAN = l + [(n/2 - cf )/f ] ×h

Where,

l = lower limit of the median class

n = number of observations  

cf = cumulative frequency  of class interval preceding the  median class

f = frequency  of median class

h = class  size

Answer 2

Answer:

Mean = ∑ f x N = 320 50 = 62.4 ∑fxN=32050=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 + n 2 − f f 1 × h =I+n2−ff1×h = 60 + 25 − 24 12 × 20 =60+25−2412×20 = 60 + 1.67 = 61.67 =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 + ( f − f 0 2 f − f 0 − f 2 ) h =I+(f−f02f−f0−f2)h = 60 + ( 12 − 10 24 − 10 − 6 ) 20 =60+(12−1024−10−6)20

=60 + 40/8 = 65