Question

The median of the frequency distribution pf following table is 48 Then find the value of f
Variable = 10,20,30,40,50,60,70,80
frequency=4,2,1,f,0,3,5,8

Answer 1

$\begin{tabular}{|c|c|} \cline{1-2} Variable &Frquency \\ \cline{1-2} 10 &4 \\ \cline{1-2} 20 &2\\ \cline{1-2} 30&1 \\ \cline{1-2} 40&f \\ \cline{1-2} 50 &0\\ \cline{1-2} 60 &3\\ \cline{1-2} 70 &5 \\ \cline{1-2} 80 &8 \\\cline{1-2} \end{tabular}$

And the mean of the following data is 48

We know

$\large \boxed{\mathsf{Mean = \dfrac{\Sigma xi}{\Sigma x} }}$

So,

$\begin{tabular}{|c|c|c|} \cline{1-3} Variable &Frquency&x \times i\\\cline{1-3} 10 &4&40\\ \cline{1-3}20 &2&40\\ \cline{1-3}30&1&30\\ \cline{1-3} 40&f&40f \\ \cline{1-3} 50 &0&0\\ \cline{1-3} 60 &3&180\\ \cline{1-3} 70 &5&350 \\ \cline{1-3} 80 &8&640\\ \cline{1-3} \end{tabular}$

$\begin{tabular}{|c|c|c|} \cline{1-3} Variable &Frquency&x \times i\\ \cline{1-3} 10 &4&40\\ \cline{1-3}20 &2&40\\ \cline{1-3} 30&1&30\\ \cline{1-3} 40&f&40f\\ \cline{1-3}50 &0&0\\ \cline{1-3}60 &3&180\\ \cline{1-3}70 &5&350 \\ \cline{1-3}80 &8&640 \\\cline{1-3}Total&23+f&1280+f \\\cline{1-3} \end{tabular}$

Since,

$\mathsf{Mean = \dfrac{\Sigma xi}{\Sigma f}}$

$\mathsf{Mean = \dfrac{1280+f}{23+f}}$

Since,

Mean = 48

So,

$\mathsf{48 = \dfrac{1280+f}{23+f}}$

$\mathsf{1104 + 23f = 1280+f}$

$\mathsf{ 23f - f = 1280 - 1104}$

$\mathsf{ 22f= 176}$

$\mathsf{\dfrac{176}{8} = f }$

$\large \boxed{\mathsf{f = 8}}$

Answer 2

Answer:

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Step-by-step explanation:

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