RBSE Solutions Class 9 Maths Chapter 15 Statistics Exercise 15.3

 

Question 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 %)

(i) Represent the information given above graphically by a bar graph.
(ii) Which condition is the major cause of women's ill health and death worldwide?
Answer:
(i) We can show the given data as a bar graph, where each bar's height represents the female fatality rate for a specific cause. Visualizing data this way helps in quick comparison of different causes.

(ii) The main reason for women's ill health and death around the world is reproductive health conditions. These issues affect a large number of women globally, highlighting a critical health concern.
In simple words: Reproductive health problems are the biggest cause of sickness and death for women worldwide. The graph clearly shows this as the largest bar.

🎯 Exam Tip: When creating a bar graph, ensure that all bars have the same width and the gaps between them are uniform. Label both axes clearly with units.

 

Question 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:

(i) Represent the information above by a bar graph.
(ii) In the class room, discuss what conclusions can be arrived from the graph.
Answer:
(i) We draw a bar graph by placing different sections of society on the x-axis and the number of girls per thousand boys on the y-axis. We set one unit of length along the y-axis to represent 10 girls. This helps compare the number of girls in different social groups easily.

(ii) (a) The graph shows that the number of girls per thousand boys is the same (920) in both Non SC/ST and Non-backward districts. This suggests similar gender ratios in these two social categories.
(b) The number of girls per thousand boys is highest in the Scheduled Tribe (ST) section (970). It is lowest in the Urban section (910). This indicates clear differences in gender ratios across different parts of society.
In simple words: The graph tells us that Non SC/ST and Non-backward districts have the same number of girls. Scheduled Tribe areas have the most girls, while city (Urban) areas have the fewest.

🎯 Exam Tip: When analyzing a bar graph, always look for the highest and lowest values to identify patterns and differences quickly. Pay attention to labels and units.

 

Question 3. 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:
(i) To represent the data, we draw a bar graph. We place the political parties (A, B, C, D, E, F) along the x-axis and the number of seats won along the y-axis. We choose a suitable scale, where, for example, 10 small divisions equal 10 seats. This makes it easy to compare the performance of each party.

(ii) Political Party A won the most seats, with a total of 75 seats. This indicates Party A had the strongest performance in the state assembly elections.
In simple words: Party A won the highest number of seats, which was 75.

🎯 Exam Tip: When drawing bar graphs, always ensure your chosen scale allows all data points to fit clearly on the graph without overcrowding. Double-check that all labels are readable.

 

Question 4. Construct histogram for the following frequency table:
Answer: We set up OX as the x-axis and OY as the y-axis on graph paper, with O as the starting point. We mark the class intervals on the x-axis, using a scale of 1 cm for every 5 units. The frequency is plotted on the y-axis, with 1 cm representing 4 units. This setup allows us to construct the histogram using rectangles (A, B, C, D, and E) that represent the given frequency table. Histograms show the distribution of continuous data.

In simple words: We draw a histogram by putting the number groups (class intervals) on the bottom line and how often they appear (frequency) on the side line. Each bar touches the next one.

🎯 Exam Tip: For histograms, ensure there are no gaps between the bars (unless a class interval has zero frequency). The width of each bar should match the class interval.

 

Question 5. Construct histogram for the following frequency table:
Answer: The histogram for the given frequency table is shown below. We place the class intervals on the x-axis and the frequencies on the y-axis to visualize the data distribution. This type of graph is very useful for understanding how data spreads across different ranges.

In simple words: The graph shows how often numbers fall into different groups. The taller the bar, the more numbers are in that group.

🎯 Exam Tip: When the frequency table is not provided, carefully read the values from the histogram's bars. Ensure your recreated graph maintains the correct heights and widths based on the visible data.

 

Question 6. Construct histogram for the following frequency table:

Answer: We represent the class interval along the x-axis and the frequency along the y-axis. Since the class intervals have different widths, we calculate the adjusted frequency (frequency density) for each to make the areas of the rectangles proportional to their frequencies. This ensures an accurate visual representation when class widths vary.

In simple words: When the groups (class intervals) are not all the same size, we adjust the bar heights so that the area of each bar correctly shows how many items are in that group.

🎯 Exam Tip: For histograms with unequal class widths, remember to calculate frequency density (frequency divided by class width) for each interval. The height of the bars should be proportional to this density, not just the frequency.

 

Question 7. Construct histogram for the following frequency distribution:

Answer: The given frequency distribution has gaps between class intervals (e.g., 5-9 then 10-14). To make it suitable for a histogram, we first convert it into a continuous distribution. We do this by finding the difference between the upper limit of one class and the lower limit of the next (here, \( 10 - 9 = 1 \)), dividing it by two (giving 0.5), and then subtracting this value from lower limits and adding it to upper limits. For instance, 5-9 becomes 4.5-9.5, and 10-14 becomes 9.5-14.5. This creates new, continuous class intervals:We then draw the histogram using these continuous intervals on the x-axis and frequencies on the y-axis.

In simple words: We change the number groups so they touch each other, then draw a bar for each group. This helps us see how often numbers appear in connected ranges.

🎯 Exam Tip: Always make sure your class intervals are continuous before drawing a histogram. If there are gaps, adjust the limits by subtracting and adding half the gap value to create touching bars.

 

Question 8. Construct the histogram from the following distribution:

Answer: The provided data gives mid values and frequencies, not direct class intervals. To construct a histogram, we first need to convert this into a continuous grouped frequency distribution. We find the class width by looking at the difference between consecutive mid-values (e.g., \( 14 - 8 = 6 \)). The lower and upper limits are found by subtracting and adding half of this width to each mid-value. For example, for a mid-value of 8 and a class width of 6, the class interval is \( 8 - \frac{6}{2} \) to \( 8 + \frac{6}{2} \), which is 5-11. We then create a table with the continuous class intervals:We set OX as the x-axis and OY as the y-axis. We choose a scale of 1 cm = 6 class intervals on the x-axis and 1 cm = 5 frequency units on the y-axis. We then draw the histogram using these intervals and frequencies. This shows how the data is grouped and where most values fall.

In simple words: First, we find the proper number groups from the mid-points given. Then we draw a histogram with bars that touch each other, showing how many times each group appears.

🎯 Exam Tip: When given mid-values, determine the class width by finding the difference between consecutive mid-values. This width helps in forming the continuous class intervals needed for a histogram.

 

Question 9. Construct a frequency polygon with the help of histogram for the following frequency distribution.

Answer: We take OX as the x-axis and OY as the y-axis, with O as the origin. We choose a scale of 1 cm = 5 units for the x-axis (for class intervals) and 1 cm = 1 unit for the y-axis (for frequency). After drawing the histogram, we mark the mid-point of the top of each rectangle. We then connect these mid-points with straight lines. We also extend the polygon to the mid-points of imaginary zero-frequency classes before the first and after the last actual class. This creates the required frequency polygon, P'PQRSTUVV', which effectively summarizes the distribution pattern.

In simple words: We draw a histogram first. Then, we find the middle of the top of each bar and connect these middle points with straight lines. We also connect the ends to the horizontal line at zero.

🎯 Exam Tip: To construct a frequency polygon, always connect the mid-points of the top of each bar in the histogram. Remember to extend the polygon to the mid-points of imaginary zero-frequency classes on the x-axis.

 

Question 10. Construct a frequency polygon for the following frequency distribution.

Answer: First, we draw a histogram for the given data. We mark the class intervals (2-4, 4-6, 6-8, 8-10) on the x-axis and the number of students (frequency) on the y-axis. Once the histogram rectangles are drawn, we mark the mid-points of the top of each rectangle. Finally, we connect these mid-points with straight lines, starting and ending at the mid-points of imaginary zero-frequency classes on the x-axis. This forms the required frequency polygon, ABCDEF, which smoothly represents the distribution of marks.

In simple words: After drawing the histogram, we connect the center of the top of each bar. We also connect the first and last points to the bottom line (x-axis) where the count is zero.

🎯 Exam Tip: When making a frequency polygon, ensure that you connect all mid-points of the histogram bars and also include the zero-frequency class marks on either end to close the polygon on the x-axis.

 

Question 11. Construct a frequency polygon for the following frequency distribution.

Answer: The given distribution is ungrouped, meaning values are distinct points rather than intervals. To construct the frequency polygon, we take the variate (x) values along the x-axis and their corresponding frequencies (f) along the y-axis. We set the scale as 1 cm = 5 units on the x-axis and 1 cm = 1 unit on the y-axis. We plot the points as (Variate, Frequency): (5,2), (10,6), (15,4), (20,1), (25,5), and (30,2). To properly close the polygon, we connect the first plotted point (5,2) to the origin (0,0) and the last plotted point (30,2) to an imaginary point (35,0) on the x-axis, assuming the next variate would be 35 with zero frequency. This completes the frequency polygon, showing how the frequency changes with each variate.

In simple words: For numbers that are not grouped, we plot each number and its count as a point. Then we connect these points with lines, making sure to also connect the first and last points to the x-axis.

🎯 Exam Tip: When plotting a frequency polygon for ungrouped data, plot the variate values directly on the x-axis and their frequencies on the y-axis. Remember to close the polygon by connecting to the x-axis at both ends.

 

Question 12. Construct a frequency polygon for the following frequency distribution.

Answer: To draw a frequency polygon without first drawing a histogram, we calculate the class mark (mid-point) for each class interval. The class mark is found by adding the lower and upper limits of an interval and dividing by two (e.g., for 0-10, the class mark is \( \frac{0+10}{2} = 5 \)). We create a table of class marks and their corresponding frequencies:We plot these class marks on the x-axis and their frequencies on the y-axis. For closure, we add imaginary class marks with zero frequency at the ends: (-5,0) before (5,8) and (75,0) after (65,16). We then join these points with straight lines to form the frequency polygon. This provides a clear visual trend of production data.

In simple words: To draw this graph without bars, we find the middle number for each group (class mark). We plot these middle numbers with their counts. Then we connect all the points, adding extra points on the bottom line at the start and end.

🎯 Exam Tip: When constructing a frequency polygon without a histogram, correctly calculate the class mark for each interval. Remember to include imaginary class marks with zero frequency at both ends to ensure the polygon touches the x-axis.