Ex.14.4 Q4 Statistics Solution - NCERT Maths Class 10

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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 \(146.75 \rm mm.\)

  
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