Ex.14.3 Q6 Statistics Solution - NCERT Maths Class 10

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Question

\(100\) surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters \(1 - 4\) \(4 - 7\) \(7 - 10\) \(10 - 13\) \(13 - 16\) \(16 - 19\)
Number of surnames \(6\) \(30\) \(40\) \(16\) \(4\) \(4\)

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Text Solution

What is known?

The frequency distribution of the number of letters in the English alphabets for \(100\)  surnames.

What is unknown?

The median and mean number of letters in the surnames and the modal size of the surnames

Reasoning:

We will find the mean by step-deviation method.

Mean,\(\overline x  = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)

Modal Class is the class with highest frequency

Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)

Where,

Class size,\(h\)

Lower limit of modal class,\(l\)

Frequency of modal class,\(f_1\)

Frequency of class preceding modal class,\(f_0\)

Frequency of class succeeding the modal class,\(f_2\)

Median Class is the class having Cumulative frequency \((cf)\) just greater than \( \frac n{2}\)

Median \( = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\)

Class size,\(h\)

Number of observations,\(n\)

Lower limit of median class,\(l\)

Frequency of median class,\(f\)

Cumulative frequency of class preceding median class,\(cf\)

Steps:

To find the median.

Number of letters    Frequency \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) Cumulative frequency (\(cf\))
\(1 - 4\) \(6\) \(6\)
\(4 - 7\) \(30\) \(30 +6= 36\)
\(7 - 10\) \(40\) \(36 + 40 = 76\)
\(10 - 13\) \(16\) \(76 + 16 = 92\)
\(13 - 16\) \(4\) \(92 + 4 = 96\)
\(16 - 19\) \(4\) \(96 + 4 = 100\)
Total (n)\(= 100\)

From the table, it can be observed that

\(n = 100{\rm{    }} \Rightarrow \frac{n}{2} = 50\)

Cumulative frequency \((cf)\) just greater than \(50\) is \(76,\) belonging to class \(7 – 10.\)

Therefore, median class \(=7 – 10\)

Class size\(, h = 3\)

Lower limit of median class\(, l = 7\)

Frequency of median class\(, f = 40\)

Cumulative frequency of class preceding median class\(, cf = 36\)

Median \( = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\)

\[\begin{array}{l}
 = 7 + \left( {\frac{{50 - 36}}{{40}}} \right) \times 3\\
 = 7 + \frac{{14}}{{40}} \times 3\\
 = 7 + \frac{{21}}{{20}}\\
 = 7 + 1.05\\
 = 8.05
\end{array}\]

To find the mean

Class mark, \({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)

Taking assumed mean\(, a = 11.5\)

Number of letters Number of Surnames \(f_i\) Class mark \(\mathbf{x}_{\mathbf{i}}\) \(\mathbf{d}_{\mathbf{i}}=\mathbf{x}_{\mathbf{i}}-\mathbf{a}\) \(\mathbf{u}_{i}=\frac{\mathbf{d}_{i}}{h}\) \(\mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}\)
\(1 - 4\) \(6\) \(2.5\) \(– 9\) \(– 3\) \(– 18\)
\(4 - 7\) \(30\) \(5.5\) \(– 6\) \(– 2\) \(– 60\)
\(7 - 10\) \(40\) \(8.5\) \(– 3\) \(– 1\) \(– 40\)
\(10 - 13\) \(16\) \(11.5\) \(0\) \(0\) \(0\)
\(13 - 16\) \(4\) \(14.5\) \(3\) \(1\) \(4\)
\(16 - 19\) \(4\) \(17.5\) \(6\) \(2\) \(8\)
Total \(100\)       \(-106\)

From the table, we obtain

\(\begin{align} \Sigma f_{i} u_{i} &=-106 \\ \Sigma f_{i} &=100\\ \text{Class size,} h &=3 \end{align}\)

\[\begin{align} \operatorname{Mean}, \overline{x} &=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h \\ &=11.5+\left(\frac{-106}{100}\right) \times 3 \\ &=11.5-3.18 \\ &=8.32 \end{align}\] 

To find mode

Number of letters Number of Surnames
\(1 - 4\) \(6\)
\(4 - 7\) \(30\)
\(7 - 10\) \(40\)
\(10 - 13\) \(16\)
\(13 - 16\) \(4\)
\(16 - 19\) \(4\)
 \(n\)\(= 100\)

From the table, it can be observed that the maximum class frequency is \(40,\) belonging to class interval \(7 - 10.\)

Class size (\(h\)) \(=3.\)

Modal class \(=7 - 10.\)

Lower limit of modal class (\(l\)\( =7.\)

Frequency of modal class \(f_1\)\(\) \(=40.\)

Frequency of class preceding the modal class \(f_0\) \(= 30.\)

Frequency of class succeeding the modal class, \(f_2\) \(=16.\)

\[\begin{align} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=7+\left[\frac{40-30}{2(40)-30-16}\right] \times 3 \\ &=7+\frac{10}{34} \times 3 \\ &=7+\frac{30}{34} \\ &=7.88\end{align}\]

Therefore, median and mean number of letters in surnames is \(8.05\) and \(8.32\) respectively while modal size of surnames is \(7.88.\)

  
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