# Ex.14.3 Q7 Statistics Solution - NCERT Maths Class 9

## Question

The runs scored by two teams \(A\) and \(B\) on the first \(60\) balls in a cricket match are given below:

Number of balls |
Team A |
Team B |

\(1 – 6\) | \(2\) | \(5\) |

\(7 – 12\) | \(1\) | \(6\) |

\(13 – 18\) | \(8\) | \(2\) |

\(19 – 24\) | \(9\) | \(10\) |

\(25 – 30\) | \(4\) | \(5\) |

\(31 – 36\) | \(5\) | \(6\) |

\(37 – 42\) | \(6\) | \(3\) |

\(43 – 48\) | \(10\) | \(4\) |

\(49 – 54\) | \(6\) | \(8\) |

\(55 – 60\) | \(2\) | \(10\) |

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

[Hint: First make the class intervals continuous.]

## Text Solution

**What is known?**

The runs scored by two teams \(A\) and \(B\) on the first \(60\) balls in a cricket match are given.

**What is Unknown?**

A frequency polygon to represent the data of both the teams.

**Reasoning:**

- It can be observed from the given data that the class intervals of the given data are not continuous. There is a gap of ‘\(I\)’ unit between them. So, to make the class intervals continuous, \(0.5 \) has to be added to every upper class limit and \(0.5\) has to be subtracted from the lower class limit.
- Class mark should also be found as below:

\[\text{Class}\,\text{Marks}=\frac{\left( \begin{align} & \text{Upper}\,\text{Limit+} \\ & \text{Lower}\,\text{Limit} \\ \end{align} \right)}{2}\]

**Steps:**

The data table with continuous interval and with class mark is as below:

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

\(0.5 \,– 6.5\) | \(3.5\) | \(2\) | \(5\) |

\(6.5 \,– 12.5\) | \(9.5 \) | \(1\) | \(6\) |

\(12.5\, – 18.5\) | \(15.5\) | \(8\) | \(2\) |

\(18.5\, – 24.5\) | \(21.5\) | \(9\) | \(10\) |

\(24.5 \,– 30.5\) | \(27.5\) | \(4\) | \(5\) |

\(30.5 \,– 36.5\) | \(33.5\) | \(5\) | \(6\) |

\(36.5 \,– 42.5\) | \(39.5\) | \(6\) | \(3\) |

\(42.5 \,– 48.5\) | \(45.5\) | \(10\) | \(4\) |

\(48.5\,– 54.5\) | \(51.5\) | \(6\) | \(8\) |

\(54.5 \,– 60.5\) | \(57.5\) | \(2\) | \(10\) |

The frequency polygon for the above data can be constructed by

(i) Number of balls on \(x\)-axis

(ii) Runs scored on \(y\)-axis with an approximate scale of “\(1\,\rm unit = 1\,\rm run\)” as the lowest run was at \(1\) and the highest was at \(10.\)