1.

Find the median of the following frequency distribution:

Answer»

Convers given less than frequency table into normal frequency table:

MarksNumber of StudentsCumulative frequency
0-101212
10-2022-12=1022
20-3035-22= 1335
30-4050-35=1550
40-5070-50= 2070
50-6086-70= 1686
60-7097-86= 1197
70-80104-97=7104
80-90109-104=5109
90-100115-109= 6115
\(\sum\)fi = 115

n = \(\sum\)fi = 115

\(\because\) n/2 = 11/2 = 57.5

\(\therefore\) Median class = 40 - 50

l = lower limit of median class = 40

c = cumulative frequency processing class of median class = 50

f = frequency of median class = 20

h = class - interval = 60 - 50 = 10

\(\therefore\) Median = l +\(\cfrac{\frac n2-c}f\times h\)

 = 40 + \(\frac{57.5-50}{20}\times10\) 

= 40 + 3.75

 = 43.75

Hence, median of marks of students is 43.75.



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