NCERT Solutions Class 9 Maths Statistics Ex 14.3 — Graphical Representation of Data

Understanding Graphical Data Representation

To solve statistics ex 14 3 class 9 ncert problems, we must master three primary types of graphical representation:

1. Bar Graphs: Used to represent qualitative characteristics or discrete variables. The width of each bar is uniform, and there are equal gaps between successive bars. The height represents the frequency of that particular category.

1. Histograms: Used to represent continuous grouped data. The class intervals are plotted on the horizontal x-axis and frequencies on the vertical y-axis. There are no gaps between the rectangular bars. However, when the class intervals have varying widths, we must compute an adjusted frequency to ensure the area of each rectangle represents the data proportionally.

1. Frequency Polygons: A continuous line graph created by joining the midpoints of the upper sides of histogram rectangles. If a histogram is not drawn, we can directly plot frequency polygons using class marks, calculated as:

$\text{Class Mark} = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$

How to Draw a Histogram with Varying Class Widths

1. Step 1: Analyze Class Intervals and Frequencies — Identify all the class intervals and note down their actual frequencies. Write down the class size (width) for each interval by subtracting the lower limit from the upper limit.
2. Step 2: Find the Minimum Class Size — Examine the class widths you computed. Identify the smallest width value in the entire set. This serves as your standard base width.
3. Step 3: Calculate Adjusted Frequencies — Compute the adjusted frequency for each class interval using the formula: Adjusted Frequency = (Given Class Frequency / Class Width) * Minimum Class Size. This ensures the area of the rectangles remains proportional to the data density.
4. Step 4: Plot the Adjusted Histogram — Draw the axes on a graph sheet. Plot class intervals on the x-axis and the computed adjusted frequencies on the y-axis. Construct rectangles where the width matches the respective class interval and the height matches its adjusted frequency.

Board Exam Tips & Mistakes to Avoid

Here are crucial tips to score full marks in your CBSE Class 9 practical graph questions:

• Use the Kink (Break): If your class intervals start from a non-zero number (e.g., 100-110), draw a kink or zig-zag line ($\sim$) near the origin on the horizontal axis to show that the scale does not start from zero.
• Define Your Scale: Always state the scale chosen on the top-right corner of your graph sheet (e.g., Scale: X-axis: 1 cm = 10 units, Y-axis: 1 cm = 5 units).
• Don't Mix Graphs: Remember, Histograms have no gaps between bars because the data is continuous, whereas Bar Graphs must have uniform gaps between bars.
• Adjusted Frequencies are mandatory for varying class widths. Plotting original frequencies on varying class widths is a major mistake that will result in lost marks.

Practice Questions with Solutions

• Q: A political party survey shows seats won by different parties: Party A (75), Party B (55), Party C (37), Party D (29), Party E (10), Party F (37). Draw a bar graph to represent this data. A: Step 1: Identify the variables. Here, the political parties (A, B, C, D, E, F) are categorical variables, so we will draw a Bar Graph with gaps. Step 2: Select a scale. Let 1 unit on the y-axis represent 10 seats. Step 3: On the x-axis, mark the parties at equal intervals. On the y-axis, mark the seat numbers from 0 to 80. Step 4: Draw vertical bars of uniform width for each party. Height of Bar A = 7.5 units (75 seats), Bar B = 5.5 units (55 seats), Bar C = 3.7 units (37 seats), Bar D = 2.9 units (29 seats), Bar E = 1 unit (10 seats), and Bar F = 3.7 units (37 seats). Final answer: The resulting bar graph visually shows Party A won the highest number of seats, while Party E won the fewest.
• Q: Construct a histogram to represent the following distribution of daily wages of 42 workers: - 100-150: 5 workers - 150-200: 10 workers - 200-250: 15 workers - 250-300: 8 workers - 300-350: 4 workers A: Step 1: Verify the class sizes. All intervals (100-150, 150-200, etc.) have an equal width of 50. Since the widths are uniform, no adjusted frequency is needed. Step 2: Draw a kink on the x-axis starting from 0 to 100, since our class starts from 100. Step 3: Choose a scale: X-axis: 1 cm = Rs. 50; Y-axis: 1 cm = 2 workers. Step 4: Draw continuous rectangular bars with heights corresponding directly to frequencies: 5 units high for [100-150], 10 units high for [150-200], 15 units high for [200-250], 8 units high for [250-300], and 4 units high for [300-350]. Final answer: Plotting these rectangles adjacently without any gaps produces the required uniform-width histogram.
• Q: Draw a histogram for the following unequal class intervals representing age groups of children in a park: - Ages 1-2: Frequency 5 - Ages 2-3: Frequency 10 - Ages 3-5: Frequency 12 - Ages 5-7: Frequency 8 - Ages 7-10: Frequency 6 A: Step 1: Identify the class widths of each interval: - 1-2: width = 1 - 2-3: width = 1 - 3-5: width = 2 - 5-7: width = 2 - 7-10: width = 3 Step 2: Find the minimum class size. The minimum class size is 1. Step 3: Compute adjusted frequency = (Class frequency / Class width) Minimum class size: - For 1-2: (5 / 1) 1 = 5 - For 2-3: (10 / 1) 1 = 10 - For 3-5: (12 / 2) 1 = 6 - For 5-7: (8 / 2) 1 = 4 - For 7-10: (6 / 3) 1 = 2 Step 4: Draw the histogram on your graph using the calculated adjusted frequencies (5, 10, 6, 4, 2) on the y-axis, and age intervals on the x-axis. Final answer: The adjusted histogram has rectangles of width 1 (heights 5, 10), width 2 (heights 6, 4), and width 3 (height 2).
• Q: Draw a frequency polygon for the given distribution: Class 10-20 (Frequency 5), 20-30 (Frequency 12), 30-40 (Frequency 18), 40-50 (Frequency 10). A: Step 1: Calculate the class marks for each class interval. Class Mark = (Lower Limit + Upper Limit) / 2. - For 10-20: Class Mark = (10 + 20) / 2 = 15 - For 20-30: Class Mark = (20 + 30) / 2 = 25 - For 30-40: Class Mark = (30 + 40) / 2 = 35 - For 40-50: Class Mark = (40 + 50) / 2 = 45 Step 2: Add two imaginary class intervals at both ends with zero frequency to close the polygon: 0-10 (Class mark = 5, Frequency = 0) and 50-60 (Class mark = 55, Frequency = 0). Step 3: Plot the points (Class Mark, Frequency) on the graph: (5, 0), (15, 5), (25, 12), (35, 18), (45, 10), (55, 0). Step 4: Connect all plotted points sequentially with straight line segments. Final answer: The joined straight line segments form the closed frequency polygon.

Frequently Asked Questions