Class 9 RS Aggarwal Chapter-17 Bar Graph, Histogram and Frequency Polygon Solutions - Free PDF Download
FAQs on RS Aggarwal Class 9 Solutions Chapter-17 Bar Graph, Histogram and Frequency Polygon
1. What is the correct step-by-step method to solve questions on constructing a bar graph for ungrouped data in RS Aggarwal Class 9 Chapter 17?
To solve problems requiring a bar graph, follow this method: First, draw two perpendicular axes, the x-axis (horizontal) and y-axis (vertical). Represent the data categories on the x-axis and the frequency on the y-axis. Choose a suitable scale for the y-axis. Then, draw rectangular bars of uniform width with equal spacing between them. The height of each bar should correspond to the frequency of its category as per the chosen scale.
2. How do you construct a histogram for a given continuous grouped frequency distribution as per the problems in RS Aggarwal?
The correct procedure for constructing a histogram for a continuous distribution involves these steps:
Represent the class intervals on the x-axis and frequencies on the y-axis.
Draw rectangles with the class intervals as bases.
The height of each rectangle must be proportional to the frequency of the corresponding class interval.
Since the data is continuous, no gaps should be left between consecutive rectangles.
3. What is the correct method to find the class mark for plotting a frequency polygon, and why is this step essential for an accurate solution?
The class mark is the midpoint of a class interval and is calculated using the formula: Class Mark = (Upper Limit + Lower Limit) / 2. This step is essential because a frequency polygon represents the frequency of each class by a single point. The class mark serves as this representative point for the entire interval. Using the correct class mark ensures that the polygon accurately depicts the distribution's shape and central tendency for each group, making it a critical step for a correct solution.
4. When solving problems, how can a frequency polygon be drawn without first constructing a histogram?
Yes, you can solve for a frequency polygon directly. The method is to first calculate the class marks for each class interval. Then, plot the points corresponding to (class mark, frequency) on a graph. Connect these points sequentially using straight line segments. To complete the solution, the polygon must be closed by joining the first and last points to the x-axis. This is done by taking two additional classes, one before the first class and one after the last class, both with a frequency of zero, and plotting their class marks.
5. How do you solve problems in RS Aggarwal that require a histogram for a discontinuous frequency distribution?
To solve such problems, you must first convert the discontinuous distribution into a continuous one. Find the gap between the upper limit of one class and the lower limit of the next. The adjustment factor is half of this gap. To create continuous classes, subtract the adjustment factor from all lower limits and add it to all upper limits. Once the class intervals are continuous, you can proceed with constructing the histogram using the standard method.
6. Why are the bars in a histogram drawn without gaps, unlike in a bar graph, and how does this affect the solution method?
The absence of gaps in a histogram is because it represents continuous data, where the upper limit of one class is the lower limit of the next. Bar graphs, in contrast, represent discrete, separate categories. This fundamental difference dictates the solution method: for a histogram, you must ensure your class intervals are continuous. If they are not, you must first apply an adjustment factor to close the gaps, which is a step not required for a bar graph.
7. What is the correct method to solve for a histogram when the given problem has unequal class intervals?
When solving problems with unequal class intervals, the frequencies must be adjusted before drawing the histogram. The height of the rectangles must be proportional to the adjusted frequency, not the original frequency. First, identify the minimum class width from the data. Then, calculate the Adjusted Frequency for each class using the formula: (Frequency of the class / Class width of the class) × Minimum class width. The histogram is then constructed using these adjusted frequencies for the heights of the rectangles.
8. What are the key differences a student must remember when solving problems that require a histogram versus a bar graph in Chapter 17?
The key differences in the solving method are based on the type of data:
Data Type: Bar graphs are used for discrete data (e.g., months, colours), while histograms are for continuous grouped data (e.g., height, weight).
Bars: In a bar graph, the bars have equal width and are separated by equal gaps. In a histogram, the bars are adjacent with no gaps, and their width corresponds to the class interval.
Representation: The height of a bar in a bar graph represents the value. In a histogram, the area of the bar (not just height) represents the frequency.
