Ex.14.3 Q9 Statistics Solution - NCERT Maths Class 9

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Question

surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:  

Number of letters Number of surnames
\(1 – 4\) \(6\)
\(4 – 6\) \(30\)
\(6 – 8\) \(44\)
\(8 – 12\) \(16\)
\(12 – 20\) \(4\)

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Text Solution

 

What is known?

Distribution of the number of letters in the English alphabet in the surnames.

What is Unknown?

(i) A Histogram to depict the given data.

(ii) The class interval in which the maximum number of surnames lies.

Reasoning:

 (i) It can be observed from the given data that it has class intervals of varying width.

(ii) The proportion of number of surnames per \(2\) letter interval (class interval of min class size for reference) can be made.

Number of letters Number of surnames Width of the class Length of rectangle
\(1 – 4\) \(6\) \(3\) \(\begin{align}\frac{6 \times 2}{3}=4\end{align}\)
\(4 – 6\) \(30\) \(2\) \(\begin{align}\frac{30 \times 2}{2}=30\end{align}\)
\(6 – 8\) \(44\) \(2\) \(\begin{align}\frac{44 \times 2}{2}=44\end{align}\)
\(8 – 12\) \(16\) \(4\) \(\begin{align}\frac{16 \times 2}{4}=8\end{align}\)
\(12 – 20\) \(4\) \(8\) \(\begin{align}\frac{4 \times 2}{8}=1\end{align}\)

Steps:

The proportion of number of surnames per \(2\) letter interval (class interval of min class size for reference) can be made as follows:

We will take the number of letters on \(x\)-axis and proportion of the number of surnames per \(2\) letter interval on \(y\)-axis. And choose an appropriate scale of \(1 \,\rm unit = 4\) surnames for \(y\)-axis.

The histogram can be constructed as follows:

The class interval in which the maximum number of surnames lies in the interval \(6 – 8\) and has surnames in it.

  
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