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Question

The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table: Find the median length of leaves.

Length in mmNumber of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

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Solution

Step 1. Calculate the cumulative frequency:

  • The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.
  • The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5..

We know that,

Median = l+[(n2-cf)f]×h

  • Class size, h
  • number of observations, n
  • Lower limit of median class, l
  • Frequency of median class, f
  • Cumulative frequency of class preceding median class, cf

Let's construct a continuous class data

Length in mmClass intervalNumber of leavesFrequency
118-126117.5-126.533
127-135126.5-135.553+5=8
136-144135.5-144.598+9=17F
145-153144.5-153.512f17+12=29
154-162153.5-162.5529+5=34
163-171162.5-171.5434+4=38
172-180171.5-180.5238+2=40n

From the table, it can be observed that

n=40n2=20

Cumulative frequency (cf) just greater than 20 is 29, belonging to the class 144.5-153.5

Therefore, the median class =144.5-153.5

Class size lowerh=9

Lower limit of median class, l=144.5

Frequency of median class,f=12

Cumulative frequency of class preceding median class,cf=17

Step 2. calculate the median.

Median=l+[(n2-cf)f]×h

=144.5+[(20-17)12]×9

=144.5+(312)×9

=144.5+94

=144.5+2.25

=146.75

Therefore, the median length of leaves is 146.75mm.


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