017. The length of 40 leaves of a plant are measures correctly to the nearest mm. Find the median length of the leaves. length in mm 145-153 163-171 136-144 118-126 127-135 172-180 2 154-162 5 Length No. of leaves 4 12 9 5 3
1 thought on “017. The length of 40 leaves of a plant are measures correctly to the nearest mm. Find<br />the median length of the leaves.<br />”
Answer:
Step-by-she 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.)
Converting the given table into exclusive form and preparing the cumulative frequency table, we get
We have, n=40
⇒
2
n
=20
The cumulative frequency just greater than
2
n
is 29 and the corresponding class is 144.5−153.5.
Thus, 144.5−153.5 is the median class such that
2
n
=20,l=144.5,cf=17,f=12, and h=9
Substituting these values in the formula
Median, M=l+
⎝
⎛
f
2
n
−cf
⎠
⎞
×h
M=144.5+(
12
20−17
)×9
M=144.5+
12
3
×3=144.5+2.25=146.75
Hence, median length =146.75 hourstep explanation:
Answer:
Step-by-she 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.)
Converting the given table into exclusive form and preparing the cumulative frequency table, we get
We have, n=40
⇒
2
n
=20
The cumulative frequency just greater than
2
n
is 29 and the corresponding class is 144.5−153.5.
Thus, 144.5−153.5 is the median class such that
2
n
=20,l=144.5,cf=17,f=12, and h=9
Substituting these values in the formula
Median, M=l+
⎝
⎛
f
2
n
−cf
⎠
⎞
×h
M=144.5+(
12
20−17
)×9
M=144.5+
12
3
×3=144.5+2.25=146.75
Hence, median length =146.75 hourstep explanation: