Ex.14.2 Q3 Statistics Solution - NCERT Maths Class 9

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Question

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

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

  1. Construct a grouped frequency distribution table with classes \(84 - 86, 86–88,\) etc.
  2. Which month or season do you think this data is about?
  3. 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.

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

  
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