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STATISTICS - Q&A

Exercise 12.1

1. 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 %):

S.No.	Causes	Female Fatality Rate (%)
1.	Reproductive health conditions	31.8
2.	Neuro-psychiatric conditions	25.4
3.	Injuries	12.4
4.	Cardiovascular conditions	4.3
5.	Respiratory conditions	4.1
6.	Other causes	22.0

(i) Represent the information given above graphically.

To represent this data graphically, a bar graph is suitable. We will plot the 'Causes' on the x-axis and the 'Female Fatality Rate (%)' on the y-axis.

Steps:
1. Draw the x-axis and y-axis.
2. Take a suitable scale on the y-axis (e.g., 1 unit = 5%).
3. Mark the causes on the x-axis at equal intervals.
4. Draw rectangular bars of equal width for each cause with heights corresponding to their fatality rates:
- Reproductive health conditions: 31.8
- Neuro-psychiatric conditions: 25.4
- Injuries: 12.4
- Cardiovascular conditions: 4.3
- Respiratory conditions: 4.1
- Other causes: 22.0

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

By observing the table or the bar graph, the highest fatality rate is 31.8%.
This corresponds to Reproductive health conditions.
So, Reproductive health conditions 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.

Two possible factors for 'Reproductive health conditions' being the major cause are:
1. Lack of proper medical facilities and infrastructure in many regions.
2. Lack of knowledge and awareness about reproductive health and hygiene among women.

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

Section	Number of girls per thousand boys
Scheduled Caste (SC)	940
Scheduled Tribe (ST)	970
Non SC/ST	920
Backward districts	950
Non-backward districts	920
Rural	930
Urban	900

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

We can represent this data using a bar graph.

Steps:
1. Plot 'Section' on the x-axis and 'Number of girls per thousand boys' on the y-axis.
2. Choose a scale for the y-axis. Since values are between 900 and 970, we can start the y-axis from 900 or use a kink/break symbol.
3. Draw bars for each section:
- SC: 940
- ST: 970
- Non SC/ST: 920
- Backward districts: 950
- Non-backward districts: 920
- Rural: 930
- Urban: 900

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

Conclusions:
1. The number of girls per thousand boys is highest for Scheduled Tribes (ST) at 970.
2. The number is lowest for Urban areas at 900.
3. Backward districts have a higher sex ratio (950) compared to non-backward districts (920).
4. Rural areas have a better sex ratio (930) compared to urban areas (900).

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

Political Party	A	B	C	D	E	F
Seats Won	75	55	37	29	10	37

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

To draw the bar graph:
1. Take 'Political Party' on x-axis and 'Seats Won' on y-axis.
2. Scale on y-axis: 1 unit = 10 seats.
3. Draw bars corresponding to the seats won:
- A: 75
- B: 55
- C: 37
- D: 29
- E: 10
- F: 37

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

From the table and graph, Party A won 75 seats, which is the highest.
So, Party A won the maximum number of seats.

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

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

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

The given class intervals are not continuous (e.g., 126 to 127 has a gap).
To make them continuous, we subtract 0.5 from the lower limit and add 0.5 to the upper limit.
Gap = 127 - 126 = 1. So, adjustment = 1/2 = 0.5.

Modified Table:

Length (in mm)	Number of leaves
117.5 - 126.5	3
126.5 - 135.5	5
135.5 - 144.5	9
144.5 - 153.5	12
153.5 - 162.5	5
162.5 - 171.5	4
171.5 - 180.5	2

Now draw the histogram with these continuous intervals on x-axis and number of leaves on y-axis.

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

Yes, a Frequency Polygon can also be used to represent this data.

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

No, it is not correct.
Reason: The data tells us that the maximum number of leaves (12) lie in the interval 145 - 153 (or 144.5 - 153.5). It does not mean all or most of them are exactly 153 mm. They can have any length within that range.

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

Life time (in hours)	Number of lamps
300 - 400	14
400 - 500	56
500 - 600	60
600 - 700	86
700 - 800	74
800 - 900	62
900 - 1000	48

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

The class intervals are already continuous.
Take 'Life time (in hours)' on x-axis and 'Number of lamps' on y-axis.
Scale: x-axis 1 cm = 100 hours, y-axis 1 cm = 10 lamps.
Draw rectangles for each interval with height corresponding to the frequency.
- 300-400: 14
- 400-500: 56
- 500-600: 60
- 600-700: 86
- 700-800: 74
- 800-900: 62
- 900-1000: 48

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

Lamps with life time more than 700 hours are those in intervals 700-800, 800-900, and 900-1000.
Total = 74 + 62 + 48
Total = 184 lamps.

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

Section A		Section B
Marks	Frequency	Marks	Frequency
0 - 10	3	0 - 10	5
10 - 20	9	10 - 20	19
20 - 30	17	20 - 30	15
30 - 40	12	30 - 40	10
40 - 50	9	40 - 50	1

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.

To draw frequency polygons, we need the class marks (mid-points) of the class intervals.
Class Mark = (Lower Limit + Upper Limit) / 2.

For Section A:
- 0-10: Class mark 5, Freq 3
- 10-20: Class mark 15, Freq 9
- 20-30: Class mark 25, Freq 17
- 30-40: Class mark 35, Freq 12
- 40-50: Class mark 45, Freq 9

For Section B:
- 0-10: Class mark 5, Freq 5
- 10-20: Class mark 15, Freq 19
- 20-30: Class mark 25, Freq 15
- 30-40: Class mark 35, Freq 10
- 40-50: Class mark 45, Freq 1

Plot these points for both sections on the same graph and join them with straight lines.
Comparison: By observing the polygons, Section A has more students scoring higher marks (range 40-50) compared to Section B. Section B has a peak at 10-20 marks, while Section A peaks at 20-30. Overall, Section A seems to have performed better.

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

Number of balls	Team A	Team B
1 - 6	2	5
7 - 12	1	6
13 - 18	8	2
19 - 24	9	10
25 - 30	4	5
31 - 36	5	6
37 - 42	6	3
43 - 48	10	4
49 - 54	6	8
55 - 60	2	10

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

The class intervals are not continuous (1-6, 7-12).
Adjustment: 7 - 6 = 1. Half of 1 is 0.5.
Subtract 0.5 from lower limit and add 0.5 to upper limit.

New Class Intervals and Class Marks (Mid-points):
1. 0.5 - 6.5 (Mid: 3.5)
2. 6.5 - 12.5 (Mid: 9.5)
3. 12.5 - 18.5 (Mid: 15.5)
... and so on.

Plot points for Team A: (3.5, 2), (9.5, 1), (15.5, 8), (21.5, 9), (27.5, 4), (33.5, 5), (39.5, 6), (45.5, 10), (51.5, 6), (57.5, 2).
Plot points for Team B: (3.5, 5), (9.5, 6), (15.5, 2), (21.5, 10), (27.5, 5), (33.5, 6), (39.5, 3), (45.5, 4), (51.5, 8), (57.5, 10).
Join the points to form two frequency polygons on the same graph.

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

Age (in years)	Number of children
1 - 2	5
2 - 3	3
3 - 5	6
5 - 7	12
7 - 10	9
10 - 15	10
15 - 17	4

Draw a histogram to represent the data above.

The class widths are varying (1, 1, 2, 2, 3, 5, 2).
We need to calculate the adjusted frequency for the histogram so that the area of the rectangle is proportional to the frequency.
Formula: Adjusted Frequency = (Frequency / Class Width) × Min Class Width.
Minimum class width = 1.

Calculations:
1. 1-2: Width 1. Adj Freq = (5/1)*1 = 5
2. 2-3: Width 1. Adj Freq = (3/1)*1 = 3
3. 3-5: Width 2. Adj Freq = (6/2)*1 = 3
4. 5-7: Width 2. Adj Freq = (12/2)*1 = 6
5. 7-10: Width 3. Adj Freq = (9/3)*1 = 3
6. 10-15: Width 5. Adj Freq = (10/5)*1 = 2
7. 15-17: Width 2. Adj Freq = (4/2)*1 = 2

Now draw the histogram with Age on x-axis and Adjusted Frequency on y-axis.

9. 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:

Number of letters	Number of surnames
1 - 4	6
4 - 6	30
6 - 8	44
8 - 12	16
12 - 20	4

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

The class widths vary: 3, 2, 2, 4, 8.
Minimum class width = 2.
Adjusted Frequencies:
1. 1-4: Width 3. Adj = (6/3)*2 = 4
2. 4-6: Width 2. Adj = (30/2)*2 = 30
3. 6-8: Width 2. Adj = (44/2)*2 = 44
4. 8-12: Width 4. Adj = (16/4)*2 = 8
5. 12-20: Width 8. Adj = (4/8)*2 = 1

Draw the histogram using these adjusted frequencies.

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

From the table, the maximum frequency is 44.
This corresponds to the class interval 6 - 8.

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Quick Review Flashcards - Click to flip and test your knowledge!

Question

Why is graphical representation often preferred over tabular representation for data comparison?

Answer

It allows relative characteristics to be visualised at a glance and is easier to understand.

Question

What is the primary visual difference between a bar graph and a histogram regarding the space between bars?

Answer

Bar graphs have equal spacing between bars, whereas histograms have no gaps between consecutive rectangles.

Question

In a standard bar graph, what determines the height of each rectangular bar?

Answer

The value of the variable being depicted.

Question

On which axis is the variable typically represented when constructing a bar graph?

Answer

The horizontal axis (x-axis).

Question

Which type of data distribution is specifically represented using a histogram?

Answer

A grouped frequency distribution with continuous class intervals.

Question

In a histogram where all class intervals have equal width, what is the relationship between the rectangle lengths and frequencies?

Answer

The lengths of the rectangles are proportional to the frequencies.

Question

What graphical symbol is used on an axis to indicate that the data scale begins at a value other than zero?

Answer

A kink (or a break).

Question

In a histogram, the _____ of the rectangles are proportional to the corresponding frequencies.

Answer

areas

Question

Why must the lengths of rectangles be modified when drawing a histogram for class intervals of varying widths?

Answer

To ensure that the areas of the rectangles remain proportional to the frequencies.

Question

What is the formula for calculating the adjusted length of a rectangle in a histogram with varying class widths?

Answer

$\frac{\text{Frequency of the class}}{\text{Width of the class}} \times \text{Minimum class size}$

Question

When modifying histogram lengths for varying widths, what specific class size is used as the multiplier in the formula?

Answer

The minimum class size found in the dataset.

Question

How is a frequency polygon constructed from a histogram?

Answer

By joining the mid-points of the upper sides of the adjacent rectangles with line segments.

Question

What is the term for the mid-point of a class interval in statistics?

Answer

Class-mark

Question

What is the formula to calculate the class-mark of a specific interval?

Answer

$\frac{\text{Upper limit} + \text{Lower limit}}{2}$

Question

To complete a frequency polygon, one must assume an additional class interval at both ends with a frequency of _____.

Answer

zero

Question

What is the primary reason for adding zero-frequency class intervals to the start and end of a frequency polygon?

Answer

To make the area under the frequency polygon equal to the area of the histogram.

Question

How can a frequency polygon be drawn independently without first creating a histogram?

Answer

By plotting and joining the points formed by the class-marks and their corresponding frequencies.

Question

On which axis are class-marks plotted when drawing an independent frequency polygon?

Answer

The horizontal axis.

Question

Under what condition is a frequency polygon particularly useful compared to other graphs?

Answer

When the data is continuous and very large, or when comparing two different sets of data of the same nature.

Question

In the context of a bar graph, what is the role of the vertical axis (y-axis)?

Answer

It displays the values of the variable (such as frequency or cost).

Question

In a histogram representing weights, if the first class is $30.5 - 35.5$, where would the 'kink' be placed?

Answer

On the horizontal axis between the origin ($0$) and the first class start point ($30.5$).

Question

If a class interval is $140 - 150$, what is the class-mark?

Answer

$145$

Question

Concept: Uniform Bar Width

Answer

In bar graphs, bars have uniform width because only their height represents the value of the variable.

Question

If a frequency polygon starts with a class $0 - 10$, how is the preceding imaginary class-mark determined?

Answer

By extending the horizontal axis in the negative direction to find the mid-point of the interval $(-10) - 0$.

Question

When comparing the performance of two different school sections on one graph, which visual representation is most effective?

Answer

Frequency polygons

Question

Term: Bivariate Axis Representation

Answer

The method of showing variables on the x-axis and their corresponding values/frequencies on the y-axis.

Question

What happens to the area of a histogram rectangle if the class width doubles but the frequency stays the same?

Answer

The area remains the same if the length is correctly adjusted to half the original frequency height.

Question

In Exercise 12.1, if 'Reproductive health conditions' has a fatality rate of $31.8\%$, what would be the height of its bar on a scale of $1 \text{ unit} = 5\%$?

Answer

$6.36 \text{ units}$

Question

How is the width of a rectangle determined when drawing a histogram?

Answer

It is equal to the class-size of the corresponding interval.

Question

If the class intervals are not continuous (e.g. $1 - 6, 7 - 12$), what step must be taken before drawing a histogram?

Answer

The class intervals must be made continuous (e.g. $0.5 - 6.5, 6.5 - 12.5$).

Question

In the adjusted frequency formula, what does 'Width of the class' represent?

Answer

The difference between the upper limit and the lower limit of that specific class interval.

Question

For the class interval $154 - 162$, what is the class size?

Answer

$8$

Question

What determines the scale of the vertical axis in a frequency distribution graph?

Answer

The maximum frequency present in the data.

Question

True or False: In a frequency polygon, the mid-points of the rectangles are joined by curved lines.

Answer

False; they are joined by straight line segments.

Question

If the minimum class size in a varying-width dataset is $5$, and a class of width $20$ has a frequency of $12$, what is the adjusted length?

Answer

$3$

Question

In an expenditure bar graph, if one unit represents $1000$ rupees, what does a bar height of $4.5$ units represent?

Answer

$4500 \text{ rupees}$

Question

What is the class-mark of the interval $35.5 - 40.5$?

Answer

$38$

Question

Why is the width of the bars in a bar graph considered unimportant for data interpretation?

Answer

Because the value is represented solely by the height, and the bars are discrete rather than continuous.

Question

In a frequency polygon, what is the frequency value of the points added to the beginning and end?

Answer

$0$

Question

When drawing a histogram for ages $1-2, 2-3, 3-5$, what is the minimum class size?

Answer

$1$

Question

What is the class-mark for the interval $70 - 100$?

Answer

$85$

Question

In a histogram for neon lamp lifetimes, if $86$ lamps last $600-700$ hours, which axis represents the number of lamps?

Answer

The vertical axis (y-axis).

Question

If a bar graph shows 'Rural' girls at $930$ and 'Urban' girls at $910$ per $1000$ boys, which section has a higher ratio of girls?

Answer

Rural

Question

What visual property of a histogram makes it look like a 'solid figure' compared to a bar graph?

Answer

The lack of gaps between the rectangular bars.

Question

In the formula for class-mark, why do we divide the sum of limits by $2$?

Answer

To find the exact centre or average of the class interval.

Question

When plotting a frequency polygon independently, what represents the x-coordinate of each point?

Answer

The class-mark of the interval.

Question

In the leaf length example, if the maximum frequency is in the $145 - 153 \text{ mm}$ interval, is it certain that the most common length is $153 \text{ mm}$?

Answer

No, it only means the most common lengths fall somewhere within that $145 - 153 \text{ mm}$ range.

Question

Which graphical method is best for showing the 'Cost of Living Index' over a series of weeks without using rectangles?

Answer

Frequency polygon

Question

In Fig 12.5, why is the rectangle for $70 - 100$ much shorter than the one for $60 - 70$ despite having a frequency of $8$?

Answer

The width is three times larger ($30$ vs $10$), so the height must be reduced to keep the area proportional to the frequency.

Question

What is the class size of the interval $118 - 126$?

Answer

$8$ (or $9$ if considering inclusive continuous boundaries $117.5-126.5$).

Question

If a frequency polygon point is $C(155, 10)$, what does the value $10$ represent?

Answer

The frequency of the class interval associated with the class-mark $155$.

Question

To find the class succeeding the last class $190 - 200$, what would be its class-mark?

Answer

$205$

Question

In a bar graph, if the scale is $1 \text{ unit} = 10 \text{ seats}$, how many units high is a bar for a party that won $75$ seats?

Answer

$7.5 \text{ units}$

Question

The _____ of a histogram rectangle is equal to the class size, while the _____ is determined by the (adjusted) frequency.

Answer

width; height

Question

When class intervals are continuous, the upper limit of one class is _____ the lower limit of the next class.

Answer

equal to

Question

In Example 3, if the frequency for $0 - 20$ is $7$ and the minimum class size is $10$, what is the length of the rectangle?

Answer

$3.5$

Question

In a cricket match frequency polygon, what would the x-axis represent?

Answer

The number of balls (using class-marks of the intervals).

Question

True or False: A bar graph can be used to represent continuous data distributions directly without modification.

Answer

False; histograms are used for continuous data.

Question

What is the result of joining $B(145, 5)$ and $C(155, 10)$ in a frequency polygon?

Answer

A line segment representing the change in frequency between those two class-marks.

Question

What is the primary requirement for a 'suitably represented' frequency polygon?

Answer

The data must be grouped into continuous class intervals.