Ex.14.1 Q6 Statistics Solution - NCERT Maths Class 10

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Question

The table below shows the daily expenditure on food of \(25\) households in a locality.

Daily expenditure (in Rs) \(100 – 150\) \(150 – 200\) \(200 – 250\) \(250 – 300\) \(300 – 350\)
Number of households \(4\) \(5\) \(12\) \(2\) \(2\)

Find the mean daily expenditure on food by a suitable method.

Text Solution

 

What is known?

The daily expenditure on food of \(25\) households in a locality.

What is unknown?

The mean daily expenditure on food.

Reasoning:

We will solve this question by step deviation method. 

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

Steps:

We know that,

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

Class size, \(h = 50\)

Taking assumed mean, \(a = 225\)

Daily expenditure

in Rs.

Number of house holds \((f_i)\) \( x_i\) \( d_i = x_i -a \) \(\begin{align}{u_i}=\frac{ d_i}{h}\end{align}\) \(f_iu_i \)
\(100 –150\) \(4\) \(125\) \(-100\) \(-2\) \(-8\)
\(150 – 200\) \(5\) \(175\) \(-50\) \(-1\) \(-5\)
\(200 – 250\) \(12\) \(225 (a)\) \(0\) \(0\) \(0\)
\(250 – 300\) \(2\) \(275\) \(50\) \(1\) \(2\)
\(300 –350\) \(2\) \(325\) \(100\) \(2\) \(4\)
  \(\Sigma f_i=25 \)       \(\Sigma f_iu_i=-7\)

From the table, we obtain

\[\begin{array}{l}
\sum {{f_i} = 25} \\
\sum {{f_i}{u_i}}  =  - 7
\end{array}\]

\[\begin{align}{\text { Mean }(\overline{{x}})}&={{a}+\left(\frac{\sum { f_iu_i }}{\sum {f} _{i}}\right) {h}} \\ {\overline{{x}}}&={225+\left(\frac{-7}{25}\right) 50} \\ {\overline{{x}}}&={225-14} \\ {\overline{{x}}}&={211}\end{align}\]

Thus, the mean daily expenditure on food is Rs \(211\).

  
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