Ex.14.4 Q4 Statistics Solution - NCERT Maths Class 10
Question
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:
Length (in mm) | Number or leaves \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) |
\(118 – 126\) | \(3\) |
\(127 – 135\) | \(5\) |
\(136 – 144\) | \(9\) |
\(145 – 153\) | \(12\) |
\(154 – 162\) | \(5\) |
\(163 – 171\) | \(4\) |
\(172 – 180\) | \(2\) |
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to \(117.5 - 126.5, 126.5 - 135.5\;...\; 171.5 - 180.5\))
Text Solution
What is known?
The lengths of \(40\) leaves of a plant measured in millimetre.
What is unknown?
The median length of the leaves.
Reasoning:
Median Class is the class having Cumulative frequency\((cf)\) just greater than \(\frac n{2}\)
Median \( = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\)
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,\(cf\)
Steps:
Length (in mm) | Number or leaves \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) |
Cumulative frequency \(cf\) |
\(117.5 – 126.5\) | \(3\) | \(3\) |
\(126.5 – 135.5\) | \(5\) | \( 3+5=8\) |
\(135.5 - 144.5\) | \(9\) | \(8+9=17\) |
\(144.5 – 153.5\) | \(12\) | \(17 + 12 = 29\) |
\(153.5 - 162.5\) | \(5\) | \(29 + 5 = 34\) |
\(162.5 - 171.5\) | \(4\) | \(34 + 4 = 38\) |
\(171.5 – 180.5\) | \(2\) | \(38 + 2 = 40\) |
\(n=40\) |
From the table, it can be observed that
\(n = 40{\rm{ }} \Rightarrow \frac{n}{2} = 20\)
Cumulative frequency \((cf) \) just greater than \(20\) is \(29,\) belonging to class \(144.5 – 153.5\)
Therefore, median class \(=144.5 – 153.5\)
Class size\(, h = 9\)
Lower limit of median class\(, l = 144.5\)
Frequency of median class\(, f = 12\)
Cumulative frequency of class preceding median class\(, cf = 17\)
Median \( = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\)
\[\begin{array}{l}
= 144.5 + \left( {\frac{{20 - 17}}{{12}}} \right) \times 9\\
= 144.5 + \frac{3}{{12}} \times 5\\
= 144.5 + \frac{5}{4}\\
= 144.5 + 1.25\\
= 145.75
\end{array}\]
Therefore, median length of leaves is \(146.75 \rm mm.\)