NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.4

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Question 1

The following distribution gives the daily income of  \(50\) workers of a factory.

Daily income (in Rs) \(100 - 120\) \(120 - 140\) \(140 - 160\) \(160 – 180\) \(180 - 200\)
Number of workers \(12\) \(14\) \(8\) \(6\) \(10\)

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution

Video Solution

What is known?

The daily income of \(50\) workers of a factory.

What is unknown?

The less than type cumulative frequency distribution and its ogive.

Reasoning:

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

Steps:

The frequency distribution table of less than type is as follows:

Daily income (in Rs) (Upper class Limits) Cumulative Frequency
Less than \(120\) \(12\)
Less than \(140\) \(12 + 14 = 26\)
Less than \(160\) \(26 + 8 = 34\)
Less than \(180\) \(34 + 6 = 40\)
Less than \(200\) \(40 + 10 = 50\)

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

Question 2

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.

Solution

Video 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.

Question 3

The following table gives production yield per hectare of wheat of \(100\) farms of a village.

Production yield(in kg/ha) \(50\)\(55\) \(55 \)\(60\) \(8\)\(65\) \(65\)\(70\) \(70\)\(75\) \(75\)\(80\)
Number of farms \(2\) \(8\) \(12\) \(24\) \(38\) \(16\)

Change the distribution to a more than type distribution and draw ogive.

Solution

Video Solution

What is known?

Production yield per hectare of wheat of \(100\) farms of a village.

What is unknown?

The more than type distribution and its ogive.

Reasoning:

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

Steps:

The cumulative frequency distribution of more than type can be obtained as follows.

Production Yield (in kg/ha)

(Lower class limits)

Cumulative Frequency
More than or equal to \(50\) \(100\)
More than or equal to \(55\) \(100 - 2 = 98\)
More than or equal to \(60\) \(98 - 8 = 90\)
More than or equal to \(65\) \(90 - 12 = 78\)
More than or equal to \(70\) \(78 - 24 = 54\)
More than or equal to \(75\) \(54 - 38 = 16\)

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

Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.4 for FREE
Ncert Class 10 Exercise 14.4
Ncert Solutions For Class 10 Maths Chapter 14 Exercise 14.4
  
Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.4 for FREE
Ncert Class 10 Exercise 14.4
Ncert Solutions For Class 10 Maths Chapter 14 Exercise 14.4
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