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No. of balls | Team A | 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 |

43 - 48 | 10 | 4 |

49 - 54 | 6 | 8 |

55 - 60 | 2 | 10 |

Answer :

It can be observed that the class inter,zals of the given data are not continuous.
There is a gap of I in between them.

Therefore, has to be added to the upper class
limits and 0.5 has to be subtracted from the lower class limits.Also, class mark of each interval can be found by using the following formula.

\(Class mark = \frac{Upper Class Limit + Lower Class Limit}{2}\)

Continuous data with class mark of each class interval can Be represented as follows.

No. of balls | Class Mark | Team A | Team B |

0.5 - 6.5 | 3.5 | 2 | 5 |

6.5 - 12.5 | 9.5 | 1 | 6 |

12.5 - 18.5 | 15.5 | 8 | 2 |

18.5 - 24.5 | 21.5 | 9 | 10 |

24.5 - 30.5 | 27.5 | 4 | 5 |

30.5 - 36.5 | 33.5 | 5 | 6 |

36.5 - 42.5 | 39.5 | 6 | 3 |

42.5 - 48.5 | 45.5 | 10 | 4 |

48.5 - 54.5 | 51.5 | 6 | 8 |

55.5 - 60.5 | 57.5 | 2 | 10 |

By taking class marks on X-axis and runs scored on y-axis, a frequency polygon can be constructed as follows.

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