GSEB Class 9 Maths Solutions Chapter 14 Statistics Exercise 14.3

 

Question 1. A survey conducted by an organization for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %). Represent the information given above graphically. Which condition is the major cause of women's ill health and death worldwide? 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) The given information can be graphically represented using a bar graph, where the 'Causes' are shown on the x-axis and the 'Female fatality rate (%)' on the y-axis, as displayed in the image provided in the solution.
(ii) Reproductive health conditions represent the biggest cause of women's ill health and death around the world.
(iii) Two important factors contributing to neuropsychiatric conditions as a major cause include:

• Lack of proper diet: A poor diet can affect brain health and mental well-being.
• Lack of advised exercises: Physical activity helps in stress reduction and improves overall mental health.

In simple words: We can show the data on a bar graph. The biggest reason for women's health issues and deaths globally is problems with reproductive health. Poor diet and not enough exercise are two major reasons for brain and mental health problems in women.

Exam Tip: When presenting graphical data, always choose the appropriate chart type (bar graph for discrete categories, histogram for continuous data) and label axes clearly. For open-ended questions like (iii), provide specific and relevant factors.

 

Question 2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of the Indian society is given below: Represent the information above by a bar graph. In the classroom discuss what conclusions can be arrived at from the graph.

Answer:
(i) The given information can be represented using a bar graph, where 'Section' is on the x-axis and 'Number of girls per thousand boys' is on the y-axis, as shown in the solution's image.
(ii) The two conclusions that we can draw from the graph are as follows:

• (a) The number of girls, when rounded to the nearest ten per thousand boys, is highest in the scheduled tribe section of society and lowest in the urban section of society.
• (b) The number of girls per thousand boys, rounded to the nearest ten, is the same for the 'Non-SC/ST' and 'Non-backward districts' sections of society.

In simple words: The data can be shown with a bar graph. We can see that Scheduled Tribes have the most girls per thousand boys, while urban areas have the fewest. Also, the number of girls per thousand boys is identical for Non-SC/ST and Non-backward districts.

Exam Tip: When interpreting bar graphs, always look for the highest and lowest values, as well as any categories that show similar values. This helps in drawing meaningful conclusions.

 

Question 3. Given below are the seats won by different political parties in the polling outcome of 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) The polling results can be clearly shown using a bar graph, with 'Political party' on the x-axis and 'Seats won' on the y-axis, as represented in the provided solution graph.
(ii) Political party A secured the highest number of seats.In simple words: We can make a bar graph to show how many seats each party won. Party A won the most seats in the election.

Exam Tip: For election results, bar graphs are effective for quick comparisons. Identify the highest bar to determine the winner easily.

 

Question 4. The length of 40 leaves of a plant is measured correct to one millimeter, and the obtained data is represented in the following table: (i) Draw a histogram to represent the given data correct to one millimeter, and the obtained data is in the following represented table: (ii) Is there any other suitable graphical representation for the same data? (iii) Is it correct to conclude that the maximum number of leaves is 153 mm long? Why?

Answer:
(i) To represent the data using a histogram, first, we need to adjust the class intervals to make them continuous. The modified continuous distribution table is as follows:
After this, a histogram can be drawn with 'Length (in mm)' on the x-axis and 'Number of leaves' on the y-axis, using the modified continuous intervals.
(ii) Yes, a frequency polygon is another suitable graphical representation for the same data.
(iii) No, it is not correct to conclude that the maximum number of leaves are precisely 153 mm long. The maximum number of leaves (12) have lengths that fall within the class interval 145-153 mm, which means their lengths are between 144.5 mm and 153.5 mm (inclusive of 144.5 but exclusive of 153.5 after continuity adjustment). We cannot pinpoint an exact length like 153 mm as the only length for the maximum count; it represents a range.In simple words: To make a histogram, we first change the length groups to flow smoothly. A frequency polygon also works for this data. We can't say most leaves are exactly 153 mm long. Instead, most leaves are in the length group of 145-153 mm.

Exam Tip: Remember to make class intervals continuous (by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit for integer data) before drawing a histogram or frequency polygon. When interpreting intervals, avoid stating a single value if the data refers to a range.

 

Question 5. The following table gives the lifetimes of 400 neon lamps: (i) Represent the given information with the help of a histogram, (ii) How many lamps have a lifetime of more than 700 hours?

Answer:
(i) The provided information can be clearly represented using a histogram. In this histogram, the 'Life time (in hours)' should be marked on the x-axis, and the 'Number of lamps' should be marked on the y-axis. Since the class intervals are continuous (e.g., 300-400, 400-500), the bars will touch, showing the distribution of lamp lifetimes. The height of each bar will correspond to the number of lamps in that specific life-time range.
(ii) To find the number of lamps that have a lifetime exceeding 700 hours, we need to sum the frequencies for the intervals above 700 hours.
Number of lamps with lifetime \( > 700 \) hours \( = (\text{Lamps in 700-800 hours}) + (\text{Lamps in 800-900 hours}) + (\text{Lamps in 900-1000 hours}) \)
Number of lamps with lifetime \( > 700 \) hours \( = 74 + 62 + 48 \)
Number of lamps with lifetime \( > 700 \) hours \( = 184 \)
Therefore, 184 lamps possess a lifetime of more than 700 hours.In simple words: We can show the lamp lifetimes with a histogram. To find lamps lasting over 700 hours, just add up the numbers for all the groups that are more than 700 hours. That means 74 + 62 + 48 gives 184 lamps.

Exam Tip: For histograms, ensure the x-axis represents continuous intervals and the y-axis represents frequency. For questions asking about 'more than' or 'less than' a certain value, correctly sum the frequencies of the relevant intervals.

 

Question 6. The following table gives the distribution of students of two sections according to the marks obtained by them: 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:
To draw frequency polygons for both sections on the same graph, we first need to determine the class marks (midpoints) for each class interval. The modified tables for Section A and Section B are as follows:

For Section A:

For Section B:

The frequency polygons are then drawn by plotting the class marks against their respective frequencies for both sections on the same graph, as illustrated in the provided image.
From observing the graph, we can see that Section A performs better than Section B when it comes to achieving good marks. Section A's polygon generally extends more to the right, showing a higher frequency of students in higher mark ranges compared to Section B, which peaks earlier.In simple words: First, we find the middle point of each mark group for both Sections A and B. Then, we plot these middle points against how many students got those marks to make two lines on one graph. Looking at the graph, students in Section A generally get better marks than students in Section B.

Exam Tip: When drawing frequency polygons, always calculate the class mark for each interval. For comparing two datasets, drawing their polygons on the same graph allows for easy visual analysis of their distributions and performance.

 

Question 7. 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 interval continuous).

Answer:
To represent the data by frequency polygons, we first need to convert the class intervals into continuous ones and calculate their class marks (midpoints). The modified table for both teams is:
With these class marks and the runs scored for Team A and Team B, two frequency polygons can be drawn on the same graph, using 'Class marks' on the x-axis and 'Runs scored' on the y-axis, as depicted in the provided solution image. This visual representation helps compare the performance distribution of both teams over the 60 balls.In simple words: To show the data for both teams on a graph, we first need to adjust the ball ranges to be continuous and find their middle points. Then, we plot these middle points against the runs scored by each team to create two lines on the same graph. This graph helps us see how each team scored their runs.

Exam Tip: Remember to always make class intervals continuous when preparing data for frequency polygons or histograms, especially if the intervals are initially discrete. The class mark (midpoint) is crucial for plotting frequency polygons.

 

Question 8. 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:
To draw a histogram for this data, we first need to adjust the class intervals to be continuous and calculate the width of each class, and then the length of the rectangle for the histogram. The modified table, assuming a minimum class size of 1, is:
Using the 'Age (in years)' on the x-axis and 'Proportion of children per 1 mark interval' (which is the length of the rectangle) on the y-axis, a histogram can be constructed, as shown in the solution's graph. This histogram will visually represent the distribution of children across different age groups.In simple words: To draw the histogram, we first make sure the age groups are smooth and then calculate how wide each group is. We also figure out the height for each bar. Then, we use these values to draw the histogram, putting ages on the bottom and the height of the bars on the side.

Exam Tip: When class intervals have different widths, remember to calculate the "adjusted frequency" or "length of the rectangle" (Frequency / Class Width * Minimum Class Width) to ensure the areas of the bars are proportional to the frequencies, which is a key characteristic of histograms.

 

Question 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: (i) Draw a histogram to depict the given information. (ii) Write the class interval in which the maximum number of surnames lies.

Answer:
(i) To construct a histogram from the given data, we need to adjust for varying class widths by calculating the length of the rectangle for each class. Assuming a minimum class size of 2 (as indicated in the solution), the modified table is as follows:
With this modified data, a histogram can be drawn with 'Number of letters' on the x-axis and 'Frequency' on the y-axis, adjusted for the lengths of the rectangles, as seen in the graph provided in the solution.
(ii) The class interval that contains the greatest number of surnames is 6-8. This interval has 44 surnames, which is the highest frequency.In simple words: To make the histogram, we adjust the values in the table based on how wide each group is. Then, we draw the histogram using these new values. The group with the most surnames is 6-8 letters long, because 44 surnames fall into that range.

Exam Tip: Always examine the class intervals before drawing a histogram. If they are unequal, you must calculate the adjusted frequencies (or length of the rectangle) to make the histogram accurate. The tallest bar in a histogram with adjusted frequencies indicates the class with the highest frequency density, which often corresponds to the maximum number of items.