The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Find the median life.
Length(in mm): 118-126 127-135 136-144 145-153 154-162 163-171 172-180
No.of leaves: 3 5 9 12 5 4 2
Answers
Length(in mm): 118-126 127-135 136-144 145-153 154-162 163-171 172-180
No.of leaves: 3 5 9 12 5 4 2
The median life is 146.75
Given,
Given is a discontinuous function. In order to find the continuous function, we need to find the gap between the lengths.
Given,
118 - 126
127 - 135
Gap is 127 - 126 = 1
⇒ Gap/2 = 1/2 = 0.5
we need to subtract Gap/2 to the lower limits and add Gap/2 to the upper limits.
Now, we have,
Length(in mm) Length(in mm): No.of leaves: Cumulative freq
(dis - cont) (cont) (fi) cf
118 - 126 175.5 - 126.5 3 3
127 - 135 126.5 - 135.5 5 3 + 5 = 8
136 - 144 135.5 - 144.5 9 8 + 9 = 17
145 - 153 144.5 - 153.5 12 17 + 12 = 29
154 - 162 153.5 - 162.5 5 29 + 5 = 34
163 - 171 162.5 - 171.5 4 34 + 4 = 38
172 - 180 171.5 - 180.5 2 38 + 2 = 40
Median = { l + [(n/2) - cf] / f }× h
n = ∑ fi = 3+5+9+12+5+4+2 ⇒ n = 40
n/2 = 40/2 ⇒ n/2 = 20
Therefore, median class is 144.5 - 153.5
where, l = lower limit of median class = 144.5
h = class - interval = 126.5 - 117.5 = 9
cf = cumulative frequency of class before the median class = 17
f = frequency of median class = 12
Median = { l + [(n/2) - cf] / f } × h
Median = { 144.5 + [20 - 17] / 12 } × 9
= 144.5 + 3/12 × 9
= 146.75
Refer to the Attachment!!
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