# NCERT Class 9 Solutions: Statistics (Chapter 14) Exercise 14.3-Part 3 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q-7 The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Number of balls | Teams A | Team B |

2 | 5 | |

1 | 6 | |

8 | 2 | |

9 | 10 | |

4 | 5 | |

5 | 6 | |

6 | 3 | |

10 | 4 | |

6 | 8 | |

2 | 10 |

Represent the table data of both the teams on the same graph by frequency polygons.

[Hint: First make the class intervals continuous.]

Solution:

Convert to continuous frequency distribution:

Difference between lower limit of class and upper limit of preceding class:

Hence, added to each upper class limit and subtract from the lower class limits.

Class marks

Obtain distribution is given as under:

Number of balls | Class marks | Team A | Team B |

3.5 | 2 | 5 | |

9.5 | 1 | 6 | |

15.5 | 8 | 2 | |

21.5 | 9 | 10 | |

27.5 | 4 | 5 | |

33.5 | 4 | 5 | |

39.5 | 6 | 3 | |

45.5 | 10 | 4 | |

51.5 | 6 | 8 | |

57.5 | 2 | 10 |

Class mark on x-axis and number of balls on y-axis

Drawn the graph:

Q-8 A random survey of the number of children of various age groups playing in park was found as follows:

Age (in years) | Number of children |

5 | |

3 | |

6 | |

12 | |

9 | |

10 | |

4 |

Draw a histogram to represent the data above.

Solution:

As given frequency data in class size are different.

So, we calculate adjusted frequency for each class using formula:

Adjusted frequency

Here, minimum class size

Age (In year) | Frequency (number of children) | Width of classes | Adjusted Frequency |

5 | 1 | ||

3 | 1 | ||

6 | 2 | ||

12 | 2 | ||

9 | 3 | ||

10 | 5 | ||

4 | 2 |

In below chart age of children on x-axis

Proportion of children per 1 year interval on y-axis