Ex.14.1 Q3 Statistics Solution - NCERT Maths Class 10
Question
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs \(18\). Find the missing frequency \( f\).
| Daily pocket allowance (in Rs) | \(11 – 13\) | \(13 – 15\) | \(15 – 17\) | \(17 – 19\) | \(19 – 21\) | \(21 – 23\) | \(23 – 25\) |
| Number of children | \(7\) | \(6\) | \(9\) | \(13\) | \(f\) | \(5\) | \(4\) |
Text Solution
What is known?
The mean pocket allowance is Rs \(18\).
The missing frequency \(f\).
Reasoning:
We will use assumed mean method to solve this question
\[\begin{align} \operatorname{Mean,}\,\,\,(\overline{{x}})={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \end{align} \]
Solution:
We know that,
\[\text{Class mark}, x_i = \frac{{\rm Upper \;class \;limit\; + \;Lower \;class \;limit}}{2} \]
Taking assumed mean\(, a = 18\)
Steps:
| Daily pocket allowance (in Rs) | No of children \( (f_i)\) | \( X_i\) | \( d_i = x_i -a \) | \( f_id_i \) |
| \(11 – 13\) | \(7\) | \(12\) | \(-6\) | \(-42\) |
| \(13 – 15\) | \(6\) | \(14\) | \(-4\) | \(-24\) |
| \(15 – 17\) | \(9\) | \(16\) | \(-2\) | \(-18\) |
| \(17 – 19\) | \(13\) | \(18 (a)\) | \(0\) | \(0\) |
| \(19 – 21\) | \(f\) | \(20\) | \(2\) | \(2f\) |
| \(21 – 23\) | \(5\) | \(22\) | \(4\) | \(20\) |
| \(23 – 25\) | \(4\) | \(24\) | \(6\) | \(24\) |
| \(\Sigma f_i=40 +f \) | \(\Sigma f_id_i=2f-40\) |
From the table, we obtain
\(\Sigma {f}_{i} = 40 + f\)
\(\Sigma {f}_{i} {d}_{i} = 2f - 40\)
\[\begin{align} \text { Mean, }\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \\ 18 &=18+\left(\frac{-40+2 {f}}{40+f}\right) \\ 18-18 &=\frac{2 {f}-40}{40+f} \\ 2f - 40 &= 0 \\ {f} &=20 \end{align}\]
Hence, the missing frequency \(f \) = \(20\).