Graphical Representation of Statistical Data — Exercise 24

Draw a histogram to represent the following data :

Answer

Steps of construction of histogram :

1. Take 2 cm along x-axis = 10 units

2. Take 1 cm along y-axis = 4 units

3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure:

Draw a histogram to represent the following data :

Answer

Steps of construction of histogram:

1. Since, the scale on x-axis starts at 150, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 150.

2. Take 2 cm along x-axis = 50 units.

3. Take 1 cm along y-axis = 2 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure :

Construct a histogram for the following frequency distribution :

Answer

The above frequency distribution is discontinuous, to convert it into continuous frequency distribution,

$\Rightarrow \text{Adjustment factor} = \dfrac{\text{Lower limit of one class - Upper limit of previous class}}{2} \\[1em] = \dfrac{13 - 12}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for the given data is :

Steps of construction :

1. Take 2 cm along x-axis = 8 units.

2. Take 1 cm along y-axis = 10 units.

3. A kink is drawn near x-axis to show that the scale starts from 4.5 and not zero.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure:

The following table shows the number of illiterate persons in the age group (10 - 69) in a town

Draw a histogram to represent the above data.

Answer

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

$\Rightarrow \text{Adjustment factor} = \dfrac{\text{Lower limit of one class - Upper limit of previous class}}{2} \\[1em] = \dfrac{20 - 19}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for the given data is :

Steps of construction :

1. Take 2 cm along x-axis = 10 years

2. Take 1 cm along y-axis = 50 units

3. A kink is drawn near x-axis to show that the scale starts from 9.5 and not zero.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure:

Draw a histogram to represent the following data :

Draw a histogram to represent the above data.

Answer

Since, the difference between the values of any two consecutive class marks is 10 (160 - 150)

∴ subtract $\dfrac{10}{2}$ = 5, from each class mark to get the lower limit of the corresponding class interval and add 5 to each class mark to get the upper limit.

Frequency distribution table :

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 145, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 145.

2. Take 2 cm along x-axis = 10 units.

3. Take 1 cm along y-axis = 4 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure :

In a study of diabetic patients in a village, the following observations were noted :

Represent the above data by a frequency polygon.

Answer

Frequency distribution table :

Steps of construction of frequency polygon:

1. Take 2 cm along x-axis = 10 years.

2. Take 1 cm along y-axis = 2 patients.

3. Find the mid-points of class intervals.

4. Find points corresponding to given frequencies of classes and the mid-points of class intervals, and plot them.

5. Join consecutive points by line segments.

6. Join first end point with mid-point of class 0 - 10 with zero frequency and join the other end with mid point of class 70 - 80 with zero frequency.

The required frequency polygon is shown alongside.

The ages (in years) of 360 patients treated in a hospital on a particular day are given below :

Draw a histogram and a frequency polygon on the same graph to represent the above data.

Answer

Frequency distribution table :

Steps of construction of histogram :

1. Take 1 cm along x-axis = 10 years.

2. Take 1 cm along y-axis = 10 patients.

3. Construct rectangles corresponding to the above continuous frequency distribution table.

Steps of construction of frequency polygon:

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line-segments.

3. Join first end point with mid-point of class 0 - 10 with zero frequency, and join the other end point with the mid point of class 70 - 80 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

Draw a histogram and the frequency polygon from the following data:

Answer

Frequency distribution table :

Steps of construction of histogram:

1. Since, the scale on x-axis starts at 15, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 15.

2. Take 1 cm along x-axis = 5 units.

3. Take 1 cm along y-axis = 10 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

Steps of construction of frequency polygon:

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line-segments.

3. Join first end point with mid-point of class 15 - 20 with zero frequency, and join the other end point with the mid point of class 50 - 55 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

Draw a histogram for the following data:

Using this histogram, draw the frequency polygon on the same graph.

Answer

Frequency distribution table :

Steps of construction of histogram:

1. Since, the scale on x-axis starts at 560, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 560.

2. Take 1 cm along x-axis = 40 units.

3. Take 1 cm along y-axis = 50 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

Steps of construction of frequency polygon:

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line-segments.

3. Join first end point with mid-point of class (560 - 600) with zero frequency, and join the other end point with the mid point of class (840 - 880) with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

Draw an ogive to represent the following data:

Answer

The cumulative frequency distribution :

Steps of construction of ogive:

1. Take 2 cm = 50 units along x-axis.

2. Take 1 cm = 25 units along y-axis.

3. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (400, 0).

4. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (450, 16), (500, 41), (550, 81), (600, 113), (650, 131), (700, 158) and (750, 167).

5. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Draw an ogive for the following frequency distribution:

Answer

The cumulative frequency distribution :

Steps of construction of ogive:

1. Take 2 cm = 10 marks along x-axis.

2. Take 1 cm = 10 students along y-axis.

3. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (0, 0).

4. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (10, 8), (20, 23), (30, 43), (40, 50), (50, 64).

5. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Draw an ogive for the following frequency distribution :

Answer

The above frequency distribution is discontinuous, to convert it into continuous frequency distribution,

$\Rightarrow \text{Adjustment factor} = \dfrac{\text{Lower limit of one class - Upper limit of previous class}}{2} \\[1em] = \dfrac{20 - 19}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for the given data is :

Steps of construction :

1. Since, the scale on x-axis starts at 9.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 9.5.

2. Take 2 cm along x-axis = 10 units

3. Take 1 cm along y-axis = 10 units

4. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (9.5, 0).

5. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (19.5, 21), (29.5, 36), (39.5, 48), (49.5, 67), (59.5, 75).

6. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Draw an ogive for the following data:

Answer

The above frequency distribution is discontinuous, to convert it into continuous frequency distribution,

$\Rightarrow \text{Adjustment factor} = \dfrac{\text{Lower limit of one class - Upper limit of previous class}}{2} \\[1em] = \dfrac{11 - 10}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for the given data is :

Steps of construction :

1. Since, the scale on x-axis starts at 10.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 10.5.

2. Take 2 cm along x-axis = 10 units

3. Take 1 cm along y-axis = 5 units

4. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (0.5, 0).

5. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (10.5, 4), (20.5, 10), (30.5, 18), (40.5, 29), (50.5, 36), (60.5, 41).

6. Join the points marked by a free hand curve.

The required ogive is shown in the below figure: