Statistics

Given: Data on causes of illness and death among women aged 15–44 worldwide.

(i) Graphical Representation (Bar Graph):

We draw a bar graph with the causes on the X-axis and the female fatality rate (%) on the Y-axis.

- Take a suitable scale on the Y-axis, e.g., 1 unit = 5%.
- Draw bars of equal width for each cause with heights proportional to the fatality rate:

| Cause | Height of bar |
|---|---|
| Reproductive health conditions | 31.8 |
| Neuropsychiatric conditions | 25.4 |
| Injuries | 12.4 |
| Cardiovascular conditions | 4.3 |
| Respiratory conditions | 4.1 |
| Other causes | 22.0 |

All bars are drawn with equal width and gaps between them. The bar for "Reproductive health conditions" is the tallest.

(ii) Major cause:

From the bar graph (and the table), Reproductive health conditions is the major cause of women's ill health and death worldwide, with the highest fatality rate of 31.8%.

(iii) Two factors responsible:

Two major factors that play a role in reproductive health conditions being the leading cause are:
1. Lack of proper medical facilities and healthcare during pregnancy and childbirth, especially in developing countries.
2. Poor nutrition and lack of awareness about reproductive health among women, leading to complications.

Given: Number of girls per thousand boys in different sections of Indian society.

(i) Bar Graph:

- X-axis: Different sections of society.
- Y-axis: Number of girls per thousand boys.
- Scale on Y-axis: Start from 900 (using a broken axis or kink) with intervals of 10, up to 980.
- Draw bars of equal width for each section:

| Section | Bar height |
|---|---|
| SC | 940 |
| ST | 970 |
| Non SC/ST | 920 |
| Backward districts | 950 |
| Non-backward districts | 920 |
| Rural | 930 |
| Urban | 910 |

The bar for ST is the tallest and the bar for Urban is the shortest.

(ii) Conclusions from the graph:

1. The Scheduled Tribe (ST) section has the highest number of girls per thousand boys (970), which is closest to equality.
2. The Urban section has the lowest number of girls per thousand boys (910), indicating a more skewed sex ratio in urban areas, possibly due to female foeticide or preference for male children.
3. Overall, in every section, the number of girls per thousand boys is less than 1000, indicating a gender imbalance across all sections of Indian society.
4. Backward districts have a better ratio (950) than non-backward districts (920), which is a noteworthy observation.

Given: Seats won by political parties A, B, C, D, E, F.

(i) Bar Graph:

- X-axis: Political parties (A, B, C, D, E, F).
- Y-axis: Number of seats won.
- Scale: 1 unit = 10 seats.
- Draw bars of equal width with heights:

| Party | Seats (Bar height) |
|---|---|
| A | 75 |
| B | 55 |
| C | 37 |
| D | 29 |
| E | 10 |
| F | 37 |

Leave equal gaps between bars. The bar for Party A is the tallest.

(ii) Party with maximum seats:

Political Party A won the maximum number of seats, i.e., 75 seats.

Given: Length of 40 leaves with class intervals that are not continuous.

Step 1: Make class intervals continuous.

The given class intervals are discontinuous (there is a gap of 1 between consecutive classes, e.g., 126 and 127). To make them continuous, subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class:

| Length (in mm) — Continuous | 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 |

(i) Histogram:

- X-axis: Length of leaves (in mm) with continuous class intervals.
- Y-axis: Number of leaves.
- Scale: 1 unit = 2 leaves (or suitable scale).
- Draw adjacent rectangles (no gaps) with:
- Class width = 9 mm for each interval.
- Heights equal to the respective frequencies: 3, 5, 9, 12, 5, 4, 2.

The tallest bar corresponds to the class interval 144.5 – 153.5 with frequency 12.

(ii) Other suitable graphical representation:

Yes, a Frequency Polygon is another suitable graphical representation for the same data, since the data is continuous and grouped.

(iii) Is it correct to say maximum leaves are 153 mm long?

No, it is not correct. The class interval 144.5 – 153.5 has the maximum frequency (12), which means the maximum number of leaves have lengths between 144.5 mm and 153.5 mm. We cannot pinpoint that all 12 leaves are exactly 153 mm long. The data only tells us the class interval, not the exact length of each leaf.

Given: Life times of 400 neon lamps.

(i) Histogram:

- X-axis: Life time of lamps (in hours). Class intervals are already continuous.
- Y-axis: Number of lamps.
- Scale: 1 unit = 10 lamps (or suitable scale).
- Draw adjacent rectangles with no gaps:

| Class Interval | Frequency (Height of bar) |
|---|---|
| 300 – 400 | 14 |
| 400 – 500 | 56 |
| 500 – 600 | 60 |
| 600 – 700 | 86 |
| 700 – 800 | 74 |
| 800 – 900 | 62 |
| 900 – 1000 | 48 |

Each bar has width = 100 hours. The tallest bar is for 600–700 hours.

(ii) Number of lamps with life time more than 700 hours:

Lamps with life time more than 700 hours belong to the intervals: 700–800, 800–900, and 900–1000.

$$\text{Number of lamps} = 74 + 62 + 48 = \mathbf{184}$$

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

Given: Marks distribution of Section A and Section B.

Step 1: Find the mid-points (class marks) of each class interval.

$$\text{Class mark} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$

| Class Interval | Class Mark | Freq. (Section A) | Freq. (Section B) |
|---|---|---|---|
| 0 – 10 | 5 | 3 | 5 |
| 10 – 20 | 15 | 9 | 19 |
| 20 – 30 | 25 | 17 | 15 |
| 30 – 40 | 35 | 12 | 10 |
| 40 – 50 | 45 | 9 | 1 |

Step 2: Add imaginary classes at both ends with frequency 0.

- Before first class: class mark = $-5$ (class $-10$ to $0$), frequency = 0 for both.
- After last class: class mark = $55$ (class $50$ to $60$), frequency = 0 for both.

Step 3: Plot the frequency polygons.

- X-axis: Class marks (–5, 5, 15, 25, 35, 45, 55).
- Y-axis: Frequency.
- Plot points for Section A: $(−5, 0),\ (5, 3),\ (15, 9),\ (25, 17),\ (35, 12),\ (45, 9),\ (55, 0)$ and join them with straight lines.
- Plot points for Section B: $(−5, 0),\ (5, 5),\ (15, 19),\ (25, 15),\ (35, 10),\ (45, 1),\ (55, 0)$ and join them with straight lines (use a different colour or dashed line).

Comparison of performance:

- Section B has more students scoring in the lower mark range (10–20) with frequency 19, compared to Section A's 9.
- Section A has more students scoring in the higher mark ranges (20–30, 30–40, 40–50) with frequencies 17, 12, and 9 respectively, compared to Section B's 15, 10, and 1.
- Therefore, Section A has performed better overall than Section B, as more students of Section A have scored higher marks.

Given: Runs scored by Team A and Team B in overs (groups of 6 balls).

Step 1: Make class intervals continuous.

The given intervals are discontinuous (gap = 1). Subtract 0.5 from lower limit and add 0.5 to upper limit:

| Continuous Interval | Class Mark | Team A | Team B |
|---|---|---|---|
| 0.5 – 6.5 | 3.5 | 2 | 5 |
| 6.5 – 12.5 | 9.5 | 1 | 6 |
| 12.5 – 18.5 | 15.5 | 8 | 2 |
| 18.5 – 24.5 | 21.5 | 9 | 10 |
| 24.5 – 30.5 | 27.5 | 4 | 5 |
| 30.5 – 36.5 | 33.5 | 5 | 6 |
| 36.5 – 42.5 | 39.5 | 6 | 3 |
| 42.5 – 48.5 | 45.5 | 10 | 4 |
| 48.5 – 54.5 | 51.5 | 6 | 8 |
| 54.5 – 60.5 | 57.5 | 2 | 10 |

Step 2: Add imaginary classes at both ends with frequency 0.

- Before first class: class mark = $-2.5$ (interval $-5.5$ to $0.5$), frequency = 0.
- After last class: class mark = $63.5$ (interval $60.5$ to $66.5$), frequency = 0.

Step 3: Plot the frequency polygons on the same graph.

- X-axis: Class marks of number of balls.
- Y-axis: Runs scored.
- Team A points: $(-2.5,\ 0),\ (3.5,\ 2),\ (9.5,\ 1),\ (15.5,\ 8),\ (21.5,\ 9),\ (27.5,\ 4),\ (33.5,\ 5),\ (39.5,\ 6),\ (45.5,\ 10),\ (51.5,\ 6),\ (57.5,\ 2),\ (63.5,\ 0)$
- Team B points: $(-2.5,\ 0),\ (3.5,\ 5),\ (9.5,\ 6),\ (15.5,\ 2),\ (21.5,\ 10),\ (27.5,\ 5),\ (33.5,\ 6),\ (39.5,\ 3),\ (45.5,\ 4),\ (51.5,\ 8),\ (57.5,\ 10),\ (63.5,\ 0)$

Join the points for each team with straight line segments using different colours or line styles (solid for Team A, dashed for Team B). Label both polygons clearly.

Given: Number of children of various age groups. The class intervals have unequal widths, so we must use frequency density (frequency per unit class width) on the Y-axis.

Step 1: Calculate class width and frequency density.

$$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$

| Age (years) | Class Width | Frequency | Frequency Density |
|---|---|---|---|
| 1 – 2 | 1 | 5 | $5 \div 1 = 5$ |
| 2 – 3 | 1 | 3 | $3 \div 1 = 3$ |
| 3 – 5 | 2 | 6 | $6 \div 2 = 3$ |
| 5 – 7 | 2 | 12 | $12 \div 2 = 6$ |
| 7 – 10 | 3 | 9 | $9 \div 3 = 3$ |
| 10 – 15 | 5 | 10 | $10 \div 5 = 2$ |
| 15 – 17 | 2 | 4 | $4 \div 2 = 2$ |

Step 2: Draw the histogram.

- X-axis: Age (in years) — mark the points 1, 2, 3, 5, 7, 10, 15, 17.
- Y-axis: Frequency density (number of children per unit age).
- Draw adjacent rectangles with:
- Width proportional to the class width.
- Height equal to the frequency density.

The rectangle for age group 5–7 is the tallest (frequency density = 6), and the rectangles for 10–15 and 15–17 are the shortest (frequency density = 2).

Given: Frequency distribution of number of letters in 100 surnames. Class intervals have unequal widths, so we use frequency density on the Y-axis.

Step 1: Calculate class width and frequency density.

$$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$

| Number of letters | Class Width | Frequency | Frequency Density |
|---|---|---|---|
| 1 – 4 | 3 | 6 | $6 \div 3 = 2$ |
| 4 – 6 | 2 | 30 | $30 \div 2 = 15$ |
| 6 – 8 | 2 | 44 | $44 \div 2 = 22$ |
| 8 – 12 | 4 | 16 | $16 \div 4 = 4$ |
| 12 – 20 | 8 | 4 | $4 \div 8 = 0.5$ |

Step 2: Draw the histogram.

- X-axis: Number of letters — mark the points 1, 4, 6, 8, 12, 20.
- Y-axis: Frequency density (number of surnames per unit letter).
- Draw adjacent rectangles with:
- Width proportional to the class width.
- Height equal to the frequency density.

The tallest bar is for the class interval 6–8 with frequency density = 22.

(ii) Class interval with maximum number of surnames:

The class interval 6 – 8 has the maximum number of surnames, i.e., 44 surnames.