A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 - 44 (in years) worldwide, found the following figures (in %):

(i) Represent the information given above graphically.

(ii) Which condition is the major cause of women’s ill health and death worldwide?

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.

Answer

(i) Steps of construction :

1. Plot the causes on x-axis.

2. Plot female fatality rate on y-axis by taking 1 cm = 4%.

3. To represent our first Head, i.e., (RHC), we draw a rectangular bar with width 1 unit and height 31.8 units.

4. Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) From graph,

We observe that,

Reproductive health conditions is the major causes of women’s ill health & death worldwide.

(iii) Two factors which play a major role in the cause are :

1. Lack of knowledge among women about various complications and steps to be taken to overcome the difficulties.
2. Lack of medical facilities.

The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

(i) Represent the information above by a bar graph.

(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Answer

Taking different sections of the society on the x-axis and number of girls per thousand boys on the y-axis, we draw a bar graph to represent the above data.

(i) Steps of construction :

1. Plot the sections on x-axis. Since, the scale on x-axis starts at 900, a break (kink) is shown near the origin on y-axis to indicate that the graph is drawn to scale beginning at 900.

2. Plot number of girls per thousand boys at y-axis by taking 1 unit = 10 girls.

3. To represent our first Head, i.e., (SC), we draw a rectangular bar with width 1 unit and height 940 units.

4. Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) From graph,

We can observe that :

(a) The number of girls per thousand boys under Non SC/ST and non-backward districts are same as 920.

(b) Number of girls under section Scheduled Tribe (ST) per thousand boys is maximum i.e., 970.

(c) Number of girls under section Urban per thousand boys is minimum i.e., 910.

Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.

(ii) Which political party won the maximum number of seats?

Answer

Taking Political Party on the x-axis and seats won on the y-axis.

(i) Steps of construction of histogram :

1. Plot the political parties on x-axis.

2. Plot number of seats won on y-axis by taking 1 unit = 10 seats.

3. To represent our first Head, i.e., party A, we draw a rectangular bar with width 1 unit and height 75 units.

4. Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) From graph,

The maximum no. of seats won are 75 by political party A.

Hence, political party A has won the maximum number of seats.

The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table :

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]

(ii) Is there any other suitable graphical representation for the same data?

(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Answer

First we convert the discontinuous classes into continuous classes because the upper limit of the first interval does not match the lower limit of the second interval.

By formula,

Adjustment factor

= $\dfrac{1}{2}\text{[Lower limit of class - Upper limit of previous class]}$

Substituting values we get :

Adjustment factor: = $\dfrac{1}{2}[127 - 126] = \dfrac{1}{2}$ = 0.5

So, 0.5 has to be added to each upper class limits and subtracted from each lower class limits to make the class interval continuous.

Since, the class intervals are continuous, now we can construct the histogram of the given frequency table.

(i) Steps of construction of histogram :

1. Plot the classes of leaves on x-axis.

2. Plot number of leaves on y-axis by taking 1 unit = 2 leaves.

3. Since, the scale on x-axis starts at 117.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 117.5.

4. We draw rectangular bars of equal width and the lengths according to the class interval's frequencies.

(ii) Frequency polygon is another suitable graphical representation for the same data.

(iii) The maximum number of leaves lie between 144.5 mm - 153.5 mm in length, which is a range.

No, we can't conclude that the maximum leaves are 153 mm long.

The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram.

(ii) How many lamps have a life time of more than 700 hours?

Answer

(i) Steps of construction of histogram :

1. Plot the lifetime (in hours) on x-axis.

2. Plot the number of lamps on y-axis by taking 1 unit = 10 lamps.

3. Since, the scale on x-axis starts at 300, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 300.

4. To represent our first Head, i.e., (300-400) we draw a rectangular bar with width 1 unit and height 14 units.

5. Similarly, other Heads are represented without leaving a gap between two consecutive bars.

(ii) The number of neon lamps having their lifetime of more than 700 hours lies in the class intervals 700 – 800, 800 – 900 and 900 – 1000.

The corresponding frequencies when added up will be (74 + 62 + 48) = 184 lamps.

Hence, 184 lamps have a life time of more than 700 hours.

The following table gives the distribution of students of two sections according to the marks obtained by them:

Section A

Section B

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

Answer

Frequency polygons can be drawn independently without drawing histograms. This requires the midpoints of the class intervals used in the data. The mid-points are called class-marks.

Class Mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

Substituting the values we get,

Class mark = $\dfrac{10 + 0}{2} = \dfrac{10}{2}$ = 5

Similarly, proceeding in this manner, we get the following table with continuous interval and class mark:

Section A

Section B

Steps of construction of frequency polygon :

1. Take class marks on the x-axis.

2. Take frequency on the y-axis with an appropriate scale of 1 unit = 2 students.

3. We draw frequency polygon for section A by plotting the class-marks on x-axis and the frequencies on y-axis, mark the points A(5, 3), B(15, 9), C(25, 17), D(35, 12) and E(45, 9).

4. We draw a frequency polygon for section B by plotting the class-marks on x-axis and the frequencies on y-axis, marks the points U(5, 5), V(15, 19), W(25, 15), X(35,10) and Y(45, 1).

5. To complete the polygon, we assume that there is a class interval with frequency zero before the first interval i.e., -10 - 0, and one after the last interval i.e. (50, 60), with class marks -5 and 55 respectively. Mark points G(-5, 0) and Z(55, 0).

6. Join the points GABCDEZ and GUVWXYZ with the help of different line segments.

7. Frequency polygon GABCDEZ is formed for Section A.

8. Frequency polygon GUVWXYZ is formed for Section B.

From graph,

We observe that,

The performance of students of Section A is better than the performance of students of Section B, as section A shows more students securing marks between class intervals 30 – 40 and 40 – 50.

The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons. [Hint : First make the class intervals continuous.]

Answer

It can be observed from the given data that the class intervals of the given data are not continuous. There is a gap of ‘1’ unit between them. So, to make the class intervals continuous, 0.5 has to be added to every upper-class limit and 0.5 has to be subtracted from the lower-class limit.

Here upper limit is 6 and lower limit is 1 so, we add and subtract 0.5 in both the term

So, 6 + 0.5 = 6.5

and 1 - 0.5 = 0.5

By formula,

Class Mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

Now, the upper limit is 6.5 and the lower limit is 0.5.

Substitute the values and we get :

So, class mark = $\dfrac{6.5 + 0.5}{2} = \dfrac{7.0}{2}$ = 3.5

Similarly, proceeding in this manner, we get the following table with continuous interval and class mark:

Steps of construction of frequency polygon :

1. Take number of balls on x-axis using class mark values.

2. Take runs scored on y-axis by taking 1 unit = 1 run

3. The lowest run is 1 and the highest is 10.

4. We draw a frequency polygon by plotting the class-marks on x-axis and the frequencies on y-axis, marks the points M(3.5, 2), N(9.5, 1), O(15.5, 8), P(21.5, 9), Q(27.5, 4), R(33.5, 5), S(39.5, 6), T(45.5, 10), U(51.5, 6), V(57.5, 2) and K(63.5, 0) by line segments.

5. For team B, we draw a frequency polygon by plotting the class-marks on x-axis and the frequencies on y-axis, marks the points A(3.5, 5), B(9.5, 6), C(15.5, 2), D(21.5, 10), E(27.5, 5), F(33.5, 6), G(39.5, 3), H(45.5, 4), I(51.5, 8), J(57.5, 10) and K(63.5, 0) by line segments.

6. To complete the polygon, we assume that there is a class interval with frequency zero before the first class i.e. (-5.5) - 0, and one after the last class interval i.e. 60.5 - 66.5, with class marks -2.5 and 63.5 respectively. Mark points L(-2.5, 0) and K(63.5, 0).

7. Join the points LMNOPQRSTUVK and LABCDEFGHIJK with the help of different line segments.

8. Frequency polygon LMNOPQRSTUVK formed for Team A

9. Frequency polygon LABCDEFGHIJK formed for Team B.

A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Answer

From the given data, we can observe that the class intervals have varying widths. The areas of the rectangles should be proportional to the frequencies in a histogram.

For example, when the class size is 5, the length of the rectangle is 10. So when the class size is 1, the length of the rectangle will be 10/5 × 1 = 2

Steps of construction :

1. Take the age of children on x-axis, using scale 1 unit = 1 year.

2. Take proportion of children per 1 year interval on y-axis, using scale 1 unit = 1 child.

3. To represent our first Head, i.e., (1 - 2) we draw a rectangular bar with width 1 unit and length 5 units.

4. Similarly, other Heads are represented without leaving a gap between two consecutive bars, according to the table.

100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Answer

1. It can be observed from the given data that it has class intervals of varying widths.

2. The proportion of the number of surnames per 2 letters (class interval of minimum class size for reference) can be made.

(i) The length of rectangles are calculated as below:

Steps of construction :

1. Take the number of letters on the x-axis using scale 1 block = 1 letter.

2. Take the proportion of the number of surnames per every 2 letter interval on the y-axis.

3. To represent our first Head, i.e., (1 - 4) we draw a rectangular bar with width 3 units and length 4 units.

4. Similarly, other Heads are represented without leaving a gap between two consecutive bars, according to the table.

(ii) From histogram,

We observe that,

In class interval 6 - 8, 44 surnames lie.

Hence, 6 - 8 is the class interval in which the maximum number of surnames lie.