CBSE Class 9 Mathematics Graphical Distribution of Data Assignment Set 02

IX - Mathematics Assignment No-01 - Graphical Representation of Statistical Data

Question 1. Draw a bar graph of the following data

Answer: To draw a bar graph for the given data:
1. Represent the months along the horizontal axis (X-axis) at uniform intervals.
2. Represent the export of bags (in lakh) along the vertical axis (Y-axis) with an appropriate scale (e.g., \(1 \text{ unit} = 1 \text{ lakh}\)).
3. Construct rectangular bars of equal width for each month with heights corresponding to their respective values: Jan (3.5), Feb (5.7), March (8.0), April (7.5), May (6.0), and June (4.3).
In simple words: A bar graph uses rectangular bars of different heights to compare categories of data visually.

Exam Tip: Ensure equal spacing between consecutive bars and maintain a consistent bar width throughout the graph.

Question 2. Draw a bar chart for the following data

Answer: To draw the bar chart:
1. Take days of the week on the X-axis.
2. Take expenditure (in Rs) on the Y-axis using a suitable scale (e.g., \(1 \text{ cm} = 10 \text{ Rs}\)).
3. Plot bars corresponding to each day: Mon (40), Tue (55), Wed (87), Th (100), Fri (95), Sat (75), and Sun (45).
In simple words: A bar chart represents numerical values across distinct items using rectangular columns.

Exam Tip: Always label both axes clearly with units (e.g., expenditure in Rs).

Question 3. Draw the histogram of the following frequency distribution.

Answer: To construct the histogram:
1. Since the class intervals are continuous and start from 50 (not 0), use a kink (zig-zag mark) on the X-axis between 0 and 50.
2. Represent class intervals (wages) along the X-axis and number of persons along the Y-axis.
3. Draw adjacent rectangles with bases equal to the class widths (50) and heights proportional to the respective frequencies (30, 45, 60, 75, 40, 35, 15).
In simple words: A histogram is a graphical representation of a grouped frequency distribution with contiguous rectangular bars.

Exam Tip: Always use a break (kink symbol) on the X-axis if the scale does not start directly from zero.

Question 4. Construct a histogram for the following frequency distribution.

Answer: To construct the histogram:
1. Since the classes start at 0, no break on the X-axis is needed.
2. Take class intervals along the X-axis (width = 20) and number of students along the Y-axis.
3. Draw adjacent rectangles corresponding to frequencies: 14, 17, 11, 10, and 8.
In simple words: Construct contiguous bars where each bar's width is the class interval and height is the frequency.

Exam Tip: Verify that the scale chosen accommodates the highest frequency clearly.

Question 5. The time taken in seconds to solve a problem on maths by each of 25 students are as follows:
30, 33, 37, 42, 43, 46, 64, 52,
40, 38, 49, 48, 46, 50, 53, 60,
59, 58, 28, 27, 26, 20, 20, 46, 16.
(i) Construct a frequency distribution for these data, using a class-interval of 10 seconds
(ii) Draw a histogram to represent the frequency distribution.
Answer:
(i) Frequency Distribution Table:
Group the given data into class intervals of width 10 (e.g., 10-20, 20-30, etc.):

(ii) Histogram:
Construct adjacent rectangles with class intervals on the X-axis and frequencies (2, 6, 4, 7, 4, 2) on the Y-axis.
In simple words: Organize raw data into class intervals using tally marks first, then plot them as a histogram.

Exam Tip: Always double-check the sum of frequencies to ensure no data point was missed.

Question 6. Draw the histogram of the following data

Answer: Since class widths are unequal (10, 20, 15, 5, 10), we must calculate adjusted frequencies (length of rectangles) proportional to a minimum class width of 5.
Formula for adjusted frequency = \(\frac{\text{Minimum class width}}{\text{Class width}} \times \text{Frequency}\)
- For 0-10 (width 10): \(\frac{5}{10} \times 8 = 4\)
- For 10-30 (width 20): \(\frac{5}{20} \times 32 = 8\)
- For 30-45 (width 15): \(\frac{5}{15} \times 18 = 6\)
- For 45-50 (width 5): \(\frac{5}{5} \times 10 = 10\)
- For 50-60 (width 10): \(\frac{5}{10} \times 6 = 3\)
Construct the histogram using these adjusted frequencies on the Y-axis.
In simple words: When class intervals are unequal, adjust the heights of the rectangles proportionally based on the minimum class width.

Exam Tip: Never plot raw frequencies directly if class sizes vary; always compute adjusted frequencies first.

Question 7. Construct a histogram for the following data

Answer: Given class marks (\(x_i\)) with uniform class size \(h = 10\).
First, find the class intervals using the formula: Lower limit = \(x_i - \frac{h}{2}\) and Upper limit = \(x_i + \frac{h}{2}\).
- For class mark 150: \(150 - 5 = 145\) to \(150 + 5 = 155\) (Interval: 145-155)
- Similarly, intervals are: 145-155, 155-165, 165-175, 175-185, 185-195, 195-205.
Construct the histogram with these class intervals and corresponding frequencies (8, 10, 25, 12, 7, 3).
In simple words: Convert class marks into proper class intervals before drawing the histogram.

Exam Tip: Subtract and add half the class size from each class mark to determine correct boundary limits.

Question 8. The following distribution presents the number of illiterate men in a town as follows. Construct a histogram for this data

Answer: Notice that the given classes are discontinuous (10-14, 15-19).
1. Convert them into continuous classes by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.
Continuous class intervals: 9.5-14.5, 14.5-19.5, 19.5-24.5, 24.5-29.5, 29.5-34.5.
2. Construct the histogram using these continuous classes and frequencies (350, 800, 750, 550, 290).
In simple words: Convert inclusive/discontinuous class intervals into exclusive/continuous intervals by applying correction factors before plotting.

Exam Tip: Always check if the upper limit of one class matches the lower limit of the next class.

Question 9. Construct a frequency polygon for the following distribution.

Answer:
1. Convert discontinuous classes into continuous classes by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
31.5-34.5, 34.5-37.5, 37.5-40.5, 40.5-43.5, 43.5-46.5, 46.5-49.5.
2. Calculate the class marks (\(x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}\)):
Class marks: 33, 36, 39, 42, 45, 48.
3. Plot the points (Class mark, Frequency): (33, 13), (36, 12), (39, 20), (42, 19), (45, 17), (48, 15), and join them with line segments. Also include bounding points (30, 0) and (51, 0) to touch the baseline.
In simple words: A frequency polygon is formed by plotting class marks against frequencies and connecting them with straight lines.

Exam Tip: Always anchor the ends of the frequency polygon to the baseline (frequency = 0) on both sides.

Question 10. Following is the distribution of ages (in years) of two groups of teachers in a secondary school.

Represent the above data by means of frequency polygon for each group on the same axes.
Answer:
1. Find the class marks for each interval (22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5, 57.5).
2. Plot the class marks against frequencies for Group A and Group B separately on the same coordinate axes.
3. Join the points of each group with distinct line styles or colors to form two frequency polygons.
In simple words: Draw two separate frequency polygons on the same graph using different colors or line styles for comparison.

Exam Tip: Always include a clear legend indicating which line represents which group.