Mathematics
The runs scored by two teams A and B on the first 42 balls in a cricket match are given below :
No. of balls | Runs scored by Team A | Runs scored by Team B |
---|---|---|
1 - 6 | 2 | 5 |
7 - 12 | 1 | 6 |
13 - 18 | 8 | 2 |
19 - 24 | 9 | 10 |
25 - 30 | 4 | 5 |
31 - 36 | 5 | 6 |
37 - 42 | 6 | 3 |
Statistics
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Answer
The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,
Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2
=
= 0.5
Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.
Classes before adjustment | Classes after adjustment | Class mark | Runs by team A | Runs by team B |
---|---|---|---|---|
1 - 6 | 0.5 - 6.5 | 3.5 | 2 | 5 |
7 - 12 | 6.5 - 12.5 | 9.5 | 1 | 6 |
13 - 18 | 12.5 - 18.5 | 15.5 | 8 | 2 |
19 - 24 | 18.5 - 24.5 | 21.5 | 9 | 10 |
25 - 30 | 24.5 - 30.5 | 27.5 | 4 | 5 |
31 - 36 | 30.5 - 36.5 | 33.5 | 5 | 6 |
37 - 42 | 36.5 - 42.5 | 39.5 | 6 | 3 |
Steps to draw frequency polygon :
Take 1 cm on x-axis = 6 balls and 1 cm on y-axis = 1 run.
For section A :
Plot the points (3.5, 2), (9.5, 1), (15.5, 8), (21.5, 9), (27.5, 4), (33.5, 5) and (39.5, 6).
Join the consecutive points by blue line segments.
Join first end point with mid-point of class (-6.5) - 0.5 with zero frequency and join other end point with mid-point of 42.5 - 48.5.
The required frequency polygon is shown by blue line segments.
For section B :
Plot the points (3.5, 5), (9.5, 6), (15.5, 2), (21.5, 10), (27.5, 5), (33.5, 6) and (39.5, 3).
Join the consecutive points by black line segments.
Join first end point with mid-point of class (-6.5) - 0.5 with zero frequency and join other end point with mid-point of 42.5 - 48.5.
The required frequency polygon is shown by black line segments.
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