Ex.14.1 Q6 Statistics Solution - NCERT Maths Class 10
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.
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 \(211\).