Ex.14.4 Q2 STATISTICS Solution - NCERT Maths Class 10

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Question

During the medical check-up of \(35\) students of a class, their weights were recorded as follows:

Weight in (Kg) Number of students
Less than \(38\) \(0\)
Less than \(40\) \(3\)
Less than \(42\) \(5\)
Less than \(44\) \(9\)
Less than \(46\) \(14\)
Less than \(48\) \(28\)
Less than \(50\) \(32\)
Less than \(52\) \(35\)

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Text Solution

What is known?

The weight of \(35\) students of a class.

What is unknown?

The less than type ogive and median weight.

Reasoning :

The representation of cumulative frequency distribution graphically is known as a cumulative frequency curve, or an ogive.

Steps:

The given cumulative frequency distributions of less than type are

Weight in (Kg) Number of students
Less than \(38\) \(0\)
Less than \(40\) \(3\)
Less than \(42\) \(5\)
Less than \(44\) \(9\)
Less than \(46\) \(14\)
Less than \(48\) \(28\)
Less than \(50\) \(32\)
Less than 52 \(35\)

Taking upper class limits on \(x\)-axis and their respective cumulative frequencies on \(y\)-axis, its ogive can be drawn as follows.

Here, \(n = 35\)  \(\Rightarrow \frac n{2}=17.5\)

Mark the point \('A'\) whose ordinate is \(17.5\) and its \(x\)-coordinate is \(46.5.\) Therefore, median of this data is \(46.5.\)

It can be observed that the difference between two consecutive upper-class limits is \(2.\) The class marks with their respective frequencies are obtained as below

Weight (in Kg) Frequency Cumulative Frequency
Less than \(38\) 0 \(0\)
\(38\) - \(40\) \(3 - 0 = 3\) \(3\)
\(40\) - \(42\) \(5 - 3 = 2\) \(5\)
\(42\) - \(44\) \(9 - 5 = 4\) \(9\)
\(44\) - \(46\) \(14 - 9 = 5\) \(14\)
\(46\) - \(48\) \(28 - 14 = 14\) \(28\)
\(48\) - \(50\) \(32 - 28 = 4\) \(32\)
\(50\) - \(52\) \(35 - 32 = 3\) \(35\)
Total \(n=35\)  

Cumulative frequency \((cf)\) just greater than \(17.5\) is \(28\), beloging to class \(46-48.\) Therefore,median class \(=46-48\)

Class size (\(h\)) \(= 2\)

Lower class limit of median class (\(l\))\(=46\)

Frequency of median class (\(f\))\(=14\)

Cumulative frequency of class preceding median class (\(cf\)) \(=14\)

\[\begin{align}\text{Median} &= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\\ & = 46 + \left( {\frac{{17.5 - 14}}{{14}}} \right) \times 2\\ &= 46 + \frac{{3.5}}{7}\\ &= 46.5 \end{align}\]

Therefore, median of this data is \(46.5.\)

Hence, the value of median is verified.

  
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