Question

Solve this question. no spamming please. this is of 9th class​ . with correct steps .​

Answer 1

Here, we have to find the mean, median & mode of the following data :

$\boxed{\begin{array}{c|c|c|c|c|c}\bf Weight\: (in \: kg)&\sf 48&\sf 50&\sf 52 &\sf 54&\sf 58\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}\\\bf No. \: of \: players &\sf 4&\sf 3&\sf 2&\sf 2 &\sf 1\end{array}}$

❏ Firstly, let's calculate for mean.

$\boxed{\begin{array}{cccc}\bf X_i \: &\bf F_i \: &\bf F_i \times X_i \: \\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 48&\sf 4&\sf 192 \\\\\sf 50 &\sf 3&\sf 150 \\\\\sf 52 &\sf 2&\sf 104 \\\\\sf 54&\sf 2&\sf 108\\\\\sf 58 &\sf 1 &\sf 58\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\ &\sf 12& \sf 612\end{array}}$

~ Calculating the mean weight.

Formula to be used for finding Mean :-

$\star\;{\boxed{\sf{\pink{Mean = \dfrac{ \sum \: (f_i \times x_i)}{ \sum \: f_i} }}}}$

$\dag\;{\underline{\frak{Putting\:the\:values\:,}}}$

$\bf{\longrightarrow \: Mean \: = \cancel\dfrac{612}{12} } \\ \\ \ \ \longrightarrow \underline{\boxed{\bf{51 \: kg}}} \: \purple{\bigstar}$

$\therefore\:{\underline{\sf{Mean\:of\:given\:data\:is\: {\textsf{\textbf{51\:kg}}}.}}}$

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❏ Now, calculating for median.

$\boxed{\begin{array}{cccc}\bf X_i \: &\bf F_i \: &\bf Cumulative \: Frequency \: \\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 48&\sf 4&\sf 7 \\\\\sf 50 &\sf 3&\sf 4 \\\\\sf 52 &\sf 2&\sf 9 \\\\\sf 54&\sf 2&\sf 11\\\\\sf 58 &\sf 1 &\sf 12\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\ &\sf 12& \end{array}}$

• Here, N = 12, which is even.

~ Calculating the median weight.

$\bf{\longrightarrow \dfrac{1}{2} \Bigg\lgroup \bigg( \dfrac{n}{2} \: \bigg){\tiny th \: } \: term + \bigg( \dfrac{n}{2} + 1 \bigg){\tiny th \: } \: term } \\ \\ \\ \longrightarrow \: \bf \dfrac{1}{2} \Bigg\lgroup \bigg( \dfrac{12}{2} \: \bigg){\tiny th \: } \: term + \bigg( \dfrac{12}{2} + 1 \bigg) \bf{\tiny th \: } \bf \: term \\ \\ \\ \longrightarrow \bf \dfrac{1}{2} \times (6 \: {\tiny th \: } term \: + 7\: {\tiny th \: } \: term) \\ \\ \\ \longrightarrow \bf \dfrac{1}{2} \times (60 + 50) \\ \\ \\ \longrightarrow \underline{\boxed{\bf{50 \: kg}}}\: \blue{\bigstar}$

$\therefore\:{\underline{\sf{Median\:of\:given\:data\:is\: {\textsf{\textbf{50\:kg}}}.}}}$

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❏ Now, calculating for mode.

Here, we need to calculate mode by using empirical formula [as per given in the question. ]

$\dag$ Empirical formula :-

$\star\;{\boxed{\sf{\purple{Mode = 3 (Median) - 2 (Mean) }}}}$

$\dag\;{\underline{\frak{Putting\:the\:values\:,}}}$

$\bf{\longrightarrow \: 3(Median) - 2 (Mean) } \\ \\ \bf{\longrightarrow \: 3 \times 50- 2 \times 51 } \\ \\ \bf{\longrightarrow \: 150 - 102 } \\ \\ \ \ \longrightarrow \underline{\boxed{\bf{48 \: kg}}} \: \red{\bigstar}$

$\therefore\:{\underline{\sf{Mode\:of\:given\:data\:is\: {\textsf{\textbf{48\:kg}}}.}}}$