Step 1: Data Tabulation and Normalization

To construct a bar graph, we first organize the provided data into a frequency distribution table. The political parties are categorical variables, and the number of seats won represents the corresponding frequency.

Step 2: Defining the Graph Coordinates

The bar graph will be constructed on a Cartesian plane where:

• The x-axis represents the categorical variable (Political Party).
• The y-axis represents the quantitative variable (Seats Won).
• Scale: $1 \text{ unit} = 10 \text{ seats}$.

Step 3: Visual Representation (Bar Graph)

Step 4: Analytical Justification

The bar graph provides a visual comparison of the nominal data [Per the Principles of Graphical Representation in Statistics]. By keeping the width of each bar constant and varying only the height proportional to the seats won, we ensure that the visual area accurately represents the magnitude of the data, preventing cognitive bias in reading the frequency distribution.

Final Solution: The bar graph above displays the distribution of seats won, where party A holds the plurality with 75 seats and party E holds the minimum with 10 seats.

Step 1: Data Tabulation and Interpretation

To determine the number of lamps with a lifetime greater than 700 hours, we must extract the relevant frequencies from the provided distribution table. The table provides lifetime intervals in hours and the corresponding frequency ($f_i$) of lamps.

Step 2: Identifying the Relevant Range

The question asks for the count of lamps with a lifetime more than 700 hours. Based on the frequency distribution provided, this encompasses all intervals starting from 700 hours up to 1000 hours. Specifically, we include:

• $700-800$ hours: $74$ lamps
• $800-900$ hours: $62$ lamps
• $900-1000$ hours: $48$ lamps

Step 3: Calculating the Cumulative Frequency

To find the total number of lamps ($N_{>700}$), we sum the frequencies corresponding to the identified intervals [Per the Additive Property of Frequencies].

$N_{>700} = \sum f_{i}$ for all $i$ where $Interval \ge 700$

$N_{>700} = 74 + 62 + 48$

Performing the addition:

$74 + 62 = 136$

$136 + 48 = 184$

Step 4: Visualizing the Distribution

Final Solution: The total number of lamps with a lifetime of more than 700 hours is 184.

Analytical Framework: Data Interpretation

To derive meaningful conclusions from the provided dataset representing the number of girls per thousand boys across six societal sections, we must first structure the data into a tabular format for comparative analysis.

Step 1: Comparative Statistical Analysis

We analyze the variance between the maximum and minimum values to understand the skewness in sex ratio distribution.

• Maximum Value ($V_{max}$): $970$ (Scheduled Tribes)
• Minimum Value ($V_{min}$): $910$ (Urban areas)
• Range ($R$): $V_{max} - V_{min} = 970 - 910 = 60$

[Per the definition of Range in statistics, this indicates a moderate degree of disparity across the segments.]

Step 2: Deriving Sociological Conclusions

Based on the quantitative data provided, the following academic conclusions are established:

1. The Highest Favorable Ratio: The Scheduled Tribe (ST) category exhibits the highest ratio of girls per thousand boys ($970$). This suggests a socio-cultural environment where, statistically, the parity between genders is more pronounced compared to other sections.
2. Urban Disparity: The Urban sector records the lowest ratio ($910$). This counters the common assumption that higher economic development in urban centers automatically correlates with better gender ratios. This suggests potential social or economic factors—such as access to selective technologies or migration patterns—that may negatively influence the female-to-male ratio in cities.
3. Regional Disparity (Backward vs. Non-Backward): Backward districts show a higher ratio ($950$) compared to Non-Backward districts ($920$). [By Comparative Data Analysis, this indicates that the index of economic advancement (Non-Backward) is not positively correlated with the sex ratio in this context.]
4. Overall Trend: Across all categories, the number of girls per thousand boys is consistently below $1000$. This confirms a nationwide deficit in the female-to-male ratio across all analyzed sections of Indian society.

Visual Representation of Trend

Final Solution: The data reveals a systemic shortfall of girls relative to boys across all societal sections, with a maximum ratio of $970$ in Scheduled Tribes and a minimum ratio of $910$ in Urban areas, indicating that economic categorization does not guarantee a balanced sex ratio.

Step 1: Data Pre-processing and Frequency Distribution Analysis

To analyze the distribution of leaf lengths, we must first examine the continuity of the class intervals provided in the data. The given data is as follows:

[Per the definition of grouped frequency distribution, intervals must be continuous to avoid data gaps. The gap between 126 and 127 is $127 - 126 = 1$ mm. The adjustment factor is $0.5$ mm.]

Step 2: Conversion to Continuous Intervals

To make the intervals continuous, subtract $0.5$ from the lower limit and add $0.5$ to the upper limit:

• $117.5 - 126.5$
• $126.5 - 135.5$
• $135.5 - 144.5$
• $144.5 - 153.5$
• $153.5 - 162.5$
• $162.5 - 171.5$
• $171.5 - 180.5$

Step 3: Evaluating the Claim

The question asks if it is correct to conclude that the maximum number of leaves are $153$ mm long. We analyze this based on the frequency table:

The modal class (the class with the highest frequency) is $144.5 - 153.5$, which contains $12$ leaves. This means there are $12$ leaves with lengths falling within the range of $144.5$ mm to $153.5$ mm.

[By the principles of Statistics, the frequency of a class interval represents the aggregate count of observations within that range, not the frequency of a specific discrete value.]

Step 4: Logical Deduction

The assertion that the "maximum number of leaves are 153 mm long" is scientifically and statistically incorrect for the following reasons:

• Range vs. Point: The data is grouped. We only know that $12$ leaves have lengths in the interval $[144.5, 153.5]$. We do not have information regarding the individual length of every leaf.
• Data Granularity: It is possible that none of the leaves are exactly $153$ mm long; they could be $145, 146, \dots, 152$ mm.
• Interpretation Error: The value $153$ is merely the upper boundary of the class interval with the highest frequency.

Final Solution: No, it is not correct to conclude that the maximum number of leaves are 153 mm long. The frequency of 12 corresponds to the entire interval of 144.5 mm to 153.5 mm, not to the specific value of 153 mm.

Step 1: Data Tabulation and Interval Verification

To construct a histogram, we must ensure that the class intervals are continuous. The provided data for the life times of 400 neon lamps is already continuous. We prepare the frequency distribution table as follows:

Step 2: Defining Axis Coordinates

• X-axis: Represents the Life Time (hours), starting from 300 to 1000.
• Y-axis: Represents the Number of Lamps, scaling from 0 to 90 (using a scale of 1 unit = 10 lamps).
• Kink Mark: Since the X-axis starts at 300 instead of 0, a "kink" (break in the axis) is applied between the origin and 300 [Per standard graphical representation protocols].

Step 3: Graphical Representation (Histogram)

Step 4: Analytical Justification

The histogram is constructed such that the area of each rectangle is proportional to the frequency of the class interval. Since the width of all class intervals is uniform ($h = 100$), the height of each rectangle corresponds directly to the absolute frequency, satisfying the criteria for a histogram of continuous grouped data [By the Fundamental Principle of Frequency Distribution Graphs].

Final Solution: The histogram represents the distribution of 400 lamps across seven class intervals, visually confirming that the mode of the life span lies within the 600-700 hour range (highest frequency of 86).

Step 1: Data Tabulation and Representation

To determine the political party with the maximum number of seats, we organize the provided polling outcome data into a frequency distribution table. This allows for a clear comparative analysis of the quantitative values.

Step 2: Identifying the Maximum Value

We apply the principle of finding the maximum element in a finite set $S = \{75, 55, 37, 29, 10, 37\}$. By comparing the magnitudes of the elements:

• $75 > 55$
• $75 > 37$
• $75 > 29$
• $75 > 10$

The maximum value in the dataset is $75$. [By the Order Property of Real Numbers, $a > b$ if $a - b > 0$. Since $75 - 55 = 20 > 0$, $75$ is strictly greater than all other integers in the set].

Step 3: Mapping the Maximum Value to the Corresponding Attribute

Referring back to the table, the value $75$ corresponds to Political Party A. [By correspondence mapping of data entries].

Step 4: Visualizing the Distribution

Final Solution: Political Party A won the maximum number of seats, with a total of 75 seats.

Step 1: Data Organization & Frequency Table

To construct a histogram for continuous data, we represent the classes along the x-axis and the frequencies along the y-axis. Since the class intervals are already provided and are continuous, we can plot them directly.

Step 2: Adjusted Frequency for Unequal Class Widths

Because the class widths (the difference between upper and lower limits) are not uniform, we must calculate the frequency density to ensure the area of each rectangle is proportional to the frequency. [Per the rule for histograms with varying intervals: $\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Minimum Class Width}$]. Let the minimum class width be $2$.

Step 3: Visualization (Histogram)

Final Solution: The histogram represents the data distribution where the height of each rectangle is proportional to the frequency, adjusted for class width variations to maintain graphical accuracy.

Step 1: Data Analysis and Continuity Check

The provided data consists of length intervals for 40 leaves. Analyzing the class intervals: 118–126, 127–135, 136–144, etc. [Per the definition of inclusive series], we observe that the upper limit of one class is not equal to the lower limit of the next class (e.g., 126 $\neq$ 127). This is a discontinuous frequency distribution.

Step 2: Transforming the Data

To represent this data graphically using a Histogram, the data must be made continuous. We calculate the adjustment factor (half the difference between the upper limit of one class and the lower limit of the next class):

$\text{Adjustment Factor} = \frac{127 - 126}{2} = 0.5$

Subtract 0.5 from each lower limit and add 0.5 to each upper limit to establish continuity [By the Principle of Class Interval Adjustment]:

Step 3: Identification of Alternative Graphical Representations

Besides the Histogram, the most suitable graphical representation for continuous grouped frequency data is the Frequency Polygon.

• Frequency Polygon: This is constructed by plotting the midpoints of the top of each rectangle in a histogram and joining them with straight lines. It effectively shows the trend of the distribution [Per Statistical Distribution Analysis].
• Frequency Curve: A smooth version of the frequency polygon, used when the sample size is large, depicting the density of the variable over the range.

Step 4: Visualizing the Frequency Polygon

Final Solution: Yes, a Frequency Polygon is a highly suitable alternative graphical representation for this data, as it allows for the analysis of the frequency distribution trends across the continuous intervals derived from the original data.

Step 1: Data Tabulation and Adjustment

To construct a histogram for continuous data where class intervals are of varying widths, we must calculate the frequency density for each interval. The frequency density is defined as:

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

Step 2: Graphical Representation Construction

We represent the Age Group along the $x$-axis and the Frequency Density along the $y$-axis. The height of each rectangular bar corresponds to the calculated frequency density, ensuring the area of the rectangle is proportional to the frequency [Per the Principle of Proportional Areas in Statistics].

Step 3: Analytical Conclusion

By normalizing the class widths via frequency density, we have successfully represented the varying intervals. The resulting histogram correctly demonstrates that the highest concentration of children (relative to width) occurs in the 5–7 age bracket [As evidenced by the maximum frequency density of 6].

Final Solution: The histogram is constructed by plotting Age (x-axis) against Frequency Density (y-axis) using the values derived in the table above, ensuring area proportionality across unequal class intervals.

Step 1: Data Organization

To construct a bar graph, we first organize the provided categorical data into a frequency table format. Each section of society serves as the independent variable ($x$-axis), and the number of girls per thousand boys serves as the dependent variable ($y$-axis).

Step 2: Graphical Representation Protocol

We represent this data using a bar graph with the following specifications:

• $X$-axis: Represents the "Section of Society". Each category is allocated a bar of uniform width.
• $Y$-axis: Represents the "Number of Girls". We utilize a broken axis (kink) starting from 900 to ensure the variance in data is clearly visible, as all values are $\geq 900$ [Per standard cartesian scaling practice].

Step 3: Visual Construction (SVG)

Step 4: Analytical Justification

Each bar height $h_i$ is determined by the formula $h_i = k(v_i - 900)$, where $v_i$ is the value of the $i$-th category and $k$ is the scaling factor (2.5 units per integer value in this SVG). This method ensures that the relative differences between societies are mathematically preserved while maintaining graph readability [By Principle of Proportionality in Data Visualization].

Final Solution: The data has been successfully mapped to a Cartesian plane where the $x$-axis categorizes society sectors and the $y$-axis represents frequency (900-980), effectively illustrating the demographic distribution as required.

Step 1: Data Tabulation and Interval Continuity

The given data uses intervals of 6 balls (e.g., 1-6, 7-12). To construct a frequency polygon, we must first establish class intervals with a common class mark or continuous boundaries. In cricket statistics, the standard approach is to use the intervals provided and identify the midpoints (class marks) for plotting.

Step 2: Defining Coordinates for Frequency Polygon

A frequency polygon is constructed by plotting the frequency against the midpoint of each class interval. To close the polygon, we include an imaginary interval at the start (midpoint $-2.5$) and end (midpoint $63.5$) with a frequency of $0$.

Step 3: Visual Representation

Step 4: Analytical Justification

• Coordinate Geometry: The horizontal axis ($x$-axis) represents the midpoint $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$. [Per the definition of Frequency Polygons in Statistics].
• Comparison: Team A (Blue) displays higher volatility in the latter half of the innings, peaking at $10$ runs in the $43-48$ ball range, while Team B (Red) shows consistent scoring in the final overs.

Final Solution: The frequency polygons represent the performance of Team A and Team B, visually demonstrating that while Team A had a significant scoring surge in the 43-48 ball interval, Team B maintained a more aggressive scoring rate during the final 55-60 ball interval.

Step 1: Determining Class Marks

To construct a frequency polygon, we must first calculate the class marks ($x_i$) for each interval. The formula for the class mark is: $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$.

Step 2: Graphical Representation

We plot the Class Marks on the $x$-axis and Frequencies on the $y$-axis. We extend the graph to include class marks $-5$ and $55$ with a frequency of $0$ to close the polygon [Per the definition of Frequency Polygon construction].

Step 3: Comparative Analysis

By observing the two polygons:

• Section A (Blue): Shows a more consistent distribution across higher marks, with a peak performance in the 20-30 marks range.
• Section B (Red): Exhibits a very sharp peak in the 10-20 marks range but shows a significant decline as the marks increase beyond 30.

Final Solution: Section A demonstrates a better overall academic performance compared to Section B, as a higher number of students in Section A achieved marks in the upper brackets (20-50), whereas Section B is heavily concentrated in the lower marks bracket (10-20).

Given Frequency Distribution Data

Step 1: Identifying the Modal Class

To determine the class interval in which the maximum number of surnames lie, we must identify the highest frequency value within the provided data set [Per the definition of the Modal Class in grouped frequency distributions].

• Comparing frequencies: $6, 30, 44, 16, 4$.
• The maximum frequency observed is $44$.
• This value of $44$ corresponds to the class interval $6 - 8$.

Step 2: Visual Representation of Frequency Distribution

Step 3: Conclusion

By observing the frequency distribution table, the value $44$ is the highest frequency. The corresponding class interval for this frequency is $6 - 8$.

Final Solution: The class interval in which the maximum number of surnames lie is 6 – 8.

Step 1: Data Organization

To represent the given statistical data graphically, we first arrange the provided causes and their corresponding fatality percentages into a structured table for systematic analysis.

Step 2: Selection of Graphical Representation

Since the data represents categorical variables with numerical frequencies (percentages) that sum to $100\%$, a Bar Graph is the optimal mathematical tool for visualization. The x-axis will represent the 'Causes of Illness and Death,' and the y-axis will represent the 'Fatality Rate (%)' [Per the Principles of Graphical Representation in Statistics].

Step 3: Graphical Construction

Step 4: Analytical Summary

• The x-axis defines the categorical variables (Causes).
• The y-axis defines the quantitative scale (%).
• Each bar's height is directly proportional to its given percentage value [Based on the Cartesian coordinate system mapping].

Final Solution: The graphical representation of the statistical data is a bar graph where the heights of the columns correspond to the respective percentages of fatality for the listed causes of illness.

Step 1: Conversion to Continuous Class Intervals

The given data contains overlapping gaps between the upper limit of one class and the lower limit of the next (e.g., $126$ and $127$). To represent this as a histogram, we must make the intervals continuous by calculating the adjustment factor: $\frac{\text{Lower Limit of current} - \text{Upper Limit of previous}}{2} = \frac{127 - 126}{2} = 0.5$.

Step 2: Graphical Representation via Histogram

We represent the continuous intervals on the $X$-axis and the frequency on the $Y$-axis. Since the scale on the $X$-axis starts at $117.5$ instead of $0$, we use a 'kink' (zig-zag mark) on the axis.

Step 3: Analytical Justification

• X-axis: Represents the length of leaves in millimeters using the continuous class boundaries.
• Y-axis: Represents the frequency (number of leaves) with a scale of $1 \text{ unit} = 1 \text{ leaf}$.
• Continuity: By setting the boundaries to end and begin at $X.5$ values, we ensure there are no gaps between bars, fulfilling the definition of a histogram [Per standard statistical practice for grouped frequency data].

Final Solution: The histogram is constructed with continuous class intervals $[117.5, 180.5]$ and heights corresponding to frequencies $[3, 5, 9, 12, 5, 4, 2]$.

Step 1: Identifying the Context of the Major Cause

In the standard dataset provided for this survey, the "major cause" (as identified in part (ii)) is Reproductive health conditions, which accounts for $31.8\%$ of illness and death among women aged $15-44$ worldwide [Per statistical analysis of the provided survey data].

Step 2: Analytical Determination of Causative Factors

To understand why reproductive health conditions represent the leading cause of morbidity and mortality in this demographic, we must examine socio-economic and biological determinants. Two primary factors are articulated below:

• Factor 1: Lack of Access to Quality Maternal Healthcare and Family Planning
  In many global regions, especially developing nations, limited access to skilled birth attendants, emergency obstetric care, and prenatal screening contributes significantly to maternal mortality. [Ref: WHO Global Health Observatory]. When family planning services are unavailable, women face higher risks associated with early or frequent pregnancies, which physically taxes the reproductive system and increases the probability of life-threatening complications like eclampsia or hemorrhage.
• Factor 2: Socio-Economic Inequality and Gender Disparities
  Socio-economic status functions as a determinant of health outcomes. Lower literacy rates and limited economic autonomy often prevent women from seeking or affording reproductive healthcare services. [Ref: Social Determinants of Health Framework]. Furthermore, cultural barriers or systemic gender discrimination may prioritize male household members for nutrition and medical resources, leaving women in the $15-44$ age cohort more vulnerable to complications during pregnancy and childbirth.

Step 3: Visual Representation of Contributing Factors

Step 4: Academic Synthesis

The high percentage value for reproductive health conditions is not merely a medical statistic but a reflection of the intersection between clinical care systems and societal structures. Improving maternal health outcomes requires a dual approach: expanding clinical infrastructure to ensure safe pregnancies and addressing the underlying poverty and gender inequality that limits health-seeking behavior. [By Public Health Axiom: Health is a result of structural, social, and biological interactions].

Final Solution: The two major factors are: 1) Insufficient access to specialized maternal healthcare infrastructure and family planning services; 2) Socio-economic inequalities and systemic gender disparities that limit women's autonomy and ability to secure preventative care.

Step 1: Data Tabulation and Extraction

Based on the provided statistical data set representing the causes of illness and death among women (ages 15-44), we organize the categories and their corresponding percentage values into the following table for systematic comparison [Per Empirical Data Analysis]:

Step 2: Comparative Analysis

To determine the "major cause," we perform a comparative analysis of the percentages [By the principle of maximum magnitude in a finite set]:

• Compare values: $31.8 > 25.4 > 22.0 > 12.4 > 4.3 > 4.1$
• Identify the maximum value: The value $31.8\%$ is the highest percentage in the data distribution.

Step 3: Visual Representation (Bar Chart)

Step 4: Logical Conclusion

Since Reproductive health conditions account for $31.8\%$ of the total, which is the largest frequency among the given set, it constitutes the major cause of women's ill health and death worldwide [Statistical Inference].

Final Solution: The major cause of women's ill health and death worldwide is Reproductive health conditions (31.8%).