Ex.14.1 Q3 Statistics Solution - NCERT Maths Class 10

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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 \(18\).

What is unknown?

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 \(f \) = \(20\).

  
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