# Ex.14.2 Q3 Statistics Solution - NCERT Maths Class 9

## Question

The relative humidity (in \(\%\)) of a certain city for a month of \(30\) days was as follows:

\(98.1\,\,\, 98.6\,\,\, 99.2\,\,\, 90.3\,\,\, 86.5\,\,\, 95.3\,\,\, 92.9\,\,\, 96.3\,\,\, 94.2\,\,\, 95.1\,\,\, \\89.2\,\,\, 92.3\,\,\, 97.1\,\,\, 93.5\,\,\, 92.7\,\,\, 95.1\,\,\, 97.2\,\,\, 93.3\,\,\, \\95.2\,\,\, 97.3\,\,\, 96.2\,\,\, 92.1\,\,\, 84.9\,\,\, 90.2\,\,\, \\95.7\,\,\, 98.3\,\,\, 97.3\,\,\, 96.1\,\,\, 92.1\,\,\, 89\)

- Construct a grouped frequency distribution table with classes \(84 - 86, 86–88,\) etc.
- Which month or season do you think this data is about?
- What is the range of this data?

## Text Solution

**What is known?**

- The relative humidity (in \(\%\)) of a certain city over a month of \(30\) days.
- Class size is \(2\), so the class intervals will be \(84-86, 86-88, 88-90\) and so on.

**What is unknown?**

- Constructing a grouped frequency distribution table.
- The month / season that the data talks about.
- Range of this data [which is the difference of the highest and the lowest values in the data.]

**Reasoning:**

By drawing frequency distribution table we can observe data month.

Range of data \(=\) maximum value \(–\) minimum value

**Steps:**

Construct a grouped frequency distribution table with class size of \(2\)

The relation humidity (in \(\%\)) of a certain city for a month can be represented as follows:

Relative Humidity (in \(\%\)) |
Number of Days (frequency) |

\(84-86\) | \(1\) |

\(86-88\) | \(1\) |

\(88-90\) | \(2\) |

\(90-92\) | \(2\) |

\(92-94\) | \(7\) |

\(94-96\) | \(6\) |

\(96-98\) | \(7\) |

\(98-100\) | \(4\) |

Total |
\(30\) |

The following features can be observed:

- The relative humidity was \(92\%\) and above, over a period of \(24 \)days \(\begin{align}(\frac{4}{5} \text{of a month})\end{align}\)
- Since the relative humidity \(\%\) is very high, it must be a data from a month of rainy season.

\[\begin{align}\text{Range of data} &= \text{Maximum value} - \text{Minimum value}\\ &= 99.2 – 84.9\\ &= 14.3\end{align}\]