Ex.14.2 Q3 Statistics Solution - NCERT Maths Class 10

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Question

The following data gives the distribution of total monthly household expenditure of \(200\) families of a village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure.

Expenditure (in ₹)

Number of families

\(1000 – 1500\)

\(1500 – 2000\)

\(2000 – 2500\)

\(2500 – 3000\)

\(3000 – 3500\)

\(3500 – 4000\)

\(4000 – 4500\)

\(4500 – 5000\)

\(24\)

\(40\)

\(33\)

\(28\)

\(30\)

\(22\)

\(16\)

\(7\)

  

Text Solution

What is known?

The total monthly household expenditure of \(200\) families of a village.

What is unknown?

The Modal and mean monthly expenditure of the families.

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,\(f_2\)

Steps:

To find mean

We know that,

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

Class size,\(h=500\)

Taking assumed mean, \(a=2750\)

Expenditure (in Rs) No of families (\(f_i\)) \(X_i\) \(u_i = \frac {(x_i - a)} {h}\) \(f_iu_i\)
\(1000 - 1500\) \(24\) \(1250\) \(-3\) \(-72\)
\(1500 - 2000\) \(40\) \(1750\) \(-2\) \(-80\)
\(2000 - 2500\) \(33\) \(2250\) \(-1\) \(-33\)
\(2500 -3000\) \(28\) \(2750(a)\) \(0\) \(0\)
\(3000 - 3500\) \(30\) \(3250\) \(1\) \(30\)
\(3500 - 4000\) \(22\) \(3750\) \(2\) \(44\)
\(4000 - 4500\) \(16\) \(4250\) \(3\) \(48\)
\(4500 - 5000\) \(7\) \(4750\) \(4\) \(28\)
  \(\Sigma f_i= 200\)     \(\Sigma f_iu_i= -35\)

From the table,we obtain

\[\begin{array}{l}
\sum {{f_i} = 200} \\
\sum {{f_i}{u_i}}  =  - 35
\end{array}\]

\[\begin{align} {\text{ Mean}\,(\overline x )} &= a + \left( \frac{\Sigma f_iu_i}{\Sigma f_i} \right)h\\ & = 2750 + \left( {\frac{{ - 35}}{{200}}} \right) \times 500\\& = \frac{{2750 + }}{{2750 + }}( - 0.175) \times 500\\ &= 2750 + ( - 87.5)\\ &= 2662.50 \end{align}\]

To find mode

Expenditure (in ₹)

Number of families

\(1000 – 1500\)

\(1500 – 2000\)

\(2000 – 2500\)

\(2500 – 3000\)

\(3000 – 3500\)

\(3500 – 4000\)

\(4000 – 4500\)

\(4500 – 5000\)

\(24\)

\(40\)

\(33\)

\(28\)

\(30\)

\(22\)

\(16\)

\(7\)

From the table, it can be observed that the maximum class frequency is \(40,\) belonging to class interval \(1500 − 2000\)

Therefore, Modal class\(=1500 − 2000\)

Class size,\(h=500\)

Lower limit of modal class,\(l=1500\)

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

Frequency of class preceding modal class,\(f_0=24\)

Frequency of class succeeding the modal class,\(f_2=33\)

\[\begin{align} {\text {Mode}} = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\end{align}\]

\[\begin{array}{l}
 = 1500 + \left( {\frac{{40 - 24}}{{2 \times 40 - 24 - 33}}} \right) \times 500\\
 = 1500 + \left( {\frac{{16}}{{80 - 57}}} \right) \times 500\\
 = 1500 + \frac{{16}}{{23}} \times 500\\
 = 1500 + 347.83\\
 = 1847.83
\end{array}\]

The modal monthly expenditure of the families is \(₹ 1847.83\)

and the mean monthly expenditure of the families is \(₹ 2662.50\)

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