Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:x \: = \: 8 \: \: }} \\ \\ [/tex]
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Calculation of mean using Step Deviation Method :-
[tex]\begin{gathered}\boxed{\begin{array}{ccccc}\sf Class\: interval&\sf f_i&\sf Mid\:Value&\sf u_i = \dfrac{x_i - 20}{2}&\sf f_i u_i\\\frac{\quad \qquad}{}&\frac{\quad\qquad}{}&\\\sf 11-13&\sf 3&\sf 12&\sf -4&\sf -12\\\\\sf 13-15 &\sf 6&\sf 14&\sf -3 &\sf -18 \\\\\sf 15-17 &\sf 9&\sf 16&\sf - 2&\sf - 18\\\\\sf 17-19&\sf 13&\sf 18&\sf - 1 &\sf - 13\\\\\sf 19-21 &\sf x &\sf 20&\sf 0&\sf 0\\\\\sf 21-23 &\sf 5 &\sf 22&\sf 1&\sf 5\\\\\sf 23-25 &\sf 4 &\sf 24&\sf 2&\sf 8\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad\qquad}{}&\frac{\qquad \qquad}{}&\frac{ \qquad\qquad}{}\\\sf Total& \sf \sum f_{i} = 40 + x&&& \sf -48\end{array}}\end{gathered} [/tex]
From above calculations, we have
[tex]\sf \: A = 20 \\ \\ [/tex]
[tex]\sf \: h = 2 \\ \\ [/tex]
[tex]\sf \: \sum \: f_i \: = \: 40 + x\\ \\ [/tex]
[tex]\sf \: \sum \: f_i \: u_i= \: - \: 48\\ \\ [/tex]
[tex]\sf \: \overline{x} \: = \: 18 \\ \\ [/tex]
We know, Mean using Step Deviation Method is given by
[tex] \: \sf \: \overline{x} = A \: + \: \dfrac{\sum \: f_i \:u_i}{\sum \: f_i \:} \times h \: \\ \\ [/tex]
So, on substituting the values, we get
[tex]\sf \: 18 = 20 + \dfrac{ - 48}{40 + x} \times 2 \\ \\ [/tex]
[tex]\sf \: 18 - 20\: = \: - \: \dfrac{96}{40 + x} \\ \\ [/tex]
[tex]\sf \: - \: 2 \: = \: - \: \dfrac{96}{40 + x} \\ \\ [/tex]
[tex]\sf \: 1 \: = \: \dfrac{48}{40 + x} \\ \\ [/tex]
[tex]\sf \: 40 + x = 48 \\ \\ [/tex]
[tex]\sf \: x = 48 - 40\\ \\ [/tex]
[tex]\bf\implies \:x\: = \: 8 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {\large{\mathfrak{Additional\:Information}}}[/tex]
1. Mean using Short Cut Method is given by
[tex] \boxed{ \sf{ \:\: \sf \: \overline{x} = A \: + \: \dfrac{\sum \: f_i \:d_i}{\sum \: f_i \:} \: \: }}\\ \\ [/tex]
2. Mean using Direct Method is given by
[tex] \boxed{ \sf{ \:\: \sf \: \overline{x} = \dfrac{\sum \: f_i \:x_i}{\sum \: f_i \:} \: \: }}\\ \\ [/tex]
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Verified answer
Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:x \: = \: 8 \: \: }} \\ \\ [/tex]
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Calculation of mean using Step Deviation Method :-
[tex]\begin{gathered}\boxed{\begin{array}{ccccc}\sf Class\: interval&\sf f_i&\sf Mid\:Value&\sf u_i = \dfrac{x_i - 20}{2}&\sf f_i u_i\\\frac{\quad \qquad}{}&\frac{\quad\qquad}{}&\\\sf 11-13&\sf 3&\sf 12&\sf -4&\sf -12\\\\\sf 13-15 &\sf 6&\sf 14&\sf -3 &\sf -18 \\\\\sf 15-17 &\sf 9&\sf 16&\sf - 2&\sf - 18\\\\\sf 17-19&\sf 13&\sf 18&\sf - 1 &\sf - 13\\\\\sf 19-21 &\sf x &\sf 20&\sf 0&\sf 0\\\\\sf 21-23 &\sf 5 &\sf 22&\sf 1&\sf 5\\\\\sf 23-25 &\sf 4 &\sf 24&\sf 2&\sf 8\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad\qquad}{}&\frac{\qquad \qquad}{}&\frac{ \qquad\qquad}{}\\\sf Total& \sf \sum f_{i} = 40 + x&&& \sf -48\end{array}}\end{gathered} [/tex]
From above calculations, we have
[tex]\sf \: A = 20 \\ \\ [/tex]
[tex]\sf \: h = 2 \\ \\ [/tex]
[tex]\sf \: \sum \: f_i \: = \: 40 + x\\ \\ [/tex]
[tex]\sf \: \sum \: f_i \: u_i= \: - \: 48\\ \\ [/tex]
[tex]\sf \: \overline{x} \: = \: 18 \\ \\ [/tex]
We know, Mean using Step Deviation Method is given by
[tex] \: \sf \: \overline{x} = A \: + \: \dfrac{\sum \: f_i \:u_i}{\sum \: f_i \:} \times h \: \\ \\ [/tex]
So, on substituting the values, we get
[tex]\sf \: 18 = 20 + \dfrac{ - 48}{40 + x} \times 2 \\ \\ [/tex]
[tex]\sf \: 18 - 20\: = \: - \: \dfrac{96}{40 + x} \\ \\ [/tex]
[tex]\sf \: - \: 2 \: = \: - \: \dfrac{96}{40 + x} \\ \\ [/tex]
[tex]\sf \: 1 \: = \: \dfrac{48}{40 + x} \\ \\ [/tex]
[tex]\sf \: 40 + x = 48 \\ \\ [/tex]
[tex]\sf \: x = 48 - 40\\ \\ [/tex]
[tex]\bf\implies \:x\: = \: 8 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {\large{\mathfrak{Additional\:Information}}}[/tex]
1. Mean using Short Cut Method is given by
[tex] \boxed{ \sf{ \:\: \sf \: \overline{x} = A \: + \: \dfrac{\sum \: f_i \:d_i}{\sum \: f_i \:} \: \: }}\\ \\ [/tex]
2. Mean using Direct Method is given by
[tex] \boxed{ \sf{ \:\: \sf \: \overline{x} = \dfrac{\sum \: f_i \:x_i}{\sum \: f_i \:} \: \: }}\\ \\ [/tex]