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Q7:

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 intervals continuous.]

Solution :

Step 1: Data Tabulation and Interval Continuity

The given data uses intervals of 6 balls (e.g., 1-6, 7-12). To construct a frequency polygon, we must first establish class intervals with a common class mark or continuous boundaries. In cricket statistics, the standard approach is to use the intervals provided and identify the midpoints (class marks) for plotting.

Balls Midpoint ($x$) Team A Team B
1-6 3.5 2 5
7-12 9.5 1 6
13-18 15.5 8 2
19-24 21.5 9 10
25-30 27.5 4 5
31-36 33.5 5 6
37-42 39.5 6 3
43-48 45.5 10 4
49-54 51.5 6 8
55-60 57.5 2 10

Step 2: Defining Coordinates for Frequency Polygon

A frequency polygon is constructed by plotting the frequency against the midpoint of each class interval. To close the polygon, we include an imaginary interval at the start (midpoint $-2.5$) and end (midpoint $63.5$) with a frequency of $0$.

Step 3: Visual Representation

Ball Intervals (Midpoints) Runs

Step 4: Analytical Justification

  • Coordinate Geometry: The horizontal axis ($x$-axis) represents the midpoint $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$. [Per the definition of Frequency Polygons in Statistics].
  • Comparison: Team A (Blue) displays higher volatility in the latter half of the innings, peaking at $10$ runs in the $43-48$ ball range, while Team B (Red) shows consistent scoring in the final overs.

Final Solution: The frequency polygons represent the performance of Team A and Team B, visually demonstrating that while Team A had a significant scoring surge in the 43-48 ball interval, Team B maintained a more aggressive scoring rate during the final 55-60 ball interval.


More Questions from Class 9 Mathematics Statistics EXERCISE 12.1


CBSE Solutions for Class 9 Mathematics Statistics


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