Question

In a math test given to 15 students the following marks out of 100 are recorded
41 39 48 52 46 62 54 40 96 52 98 40 42 52 60 Find mean median mode

Answer 1

Answer

$\Large\frak{\underline{\underline{Correct \: Question:}}}$

In a maths test given to 15 students the following marks out of 100 are recorded:

41,39,48,52,46,62,54,40,96,52,98,40,42,52,60

Find the Mean,Median and Mode of data.

$\rule{170}2$

$\Large\frak{\underline{\underline{Your~Answer:}}}$

$\underline{\bigstar\:\textsf{Mean \: of \: data:}}$

$\maltese\:\:\:\boxed{\normalsize\sf\ Mean = \frac{Sum \: of \: all \: observations}{Number \: of \: total \: observations}}\\\\\\\normalsize\ : \implies\sf\ Mean = \scriptsize\frac{(41 +39+48+52+46+62+54+40+96+52+98+40+42+52+60)}{15}\\\\\\\normalsize\ : \implies\sf\ Mean = \frac{\cancel{822}}{\cancel{15}}\\\\\\\normalsize\ : \implies\sf\ Mean = 54.8\\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{ Mean = 54.8}}}}$

$\underline{\bigstar\:\textsf{Median \: of \: data:}}\\ \\ \normalsize\sf\bullet\: Arrange \: the \: data \: into \: ascending \: order:\\ 39,40,40,42,46,48,52,52,52,54,60,62,96,98\\ \sf\bullet\ Number \: of \: observations(n) = 15 = odd \: number \\\\\maltese\:\:\boxed{\sf\ Median = \frac{n +1}{2}^{th} \: term}\\\\\\\normalsize\ : \implies\sf\ Median = \frac{15 + 1}{2}^{th} \: term\\\\\\\normalsize\ : \implies\sf\ Median = \frac{\cancel{16}}{\cancel{2}}^{th} \: term\\\\\\\normalsize\ : \implies\sf\ Median = 8^{th} \: term\\\\\\\normalsize\ : \implies\sf\ Median = 52\\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{Median = 52 }}}}$

$\underline{\bigstar\:\textsf{Mode \: of \: data:}}\\\\ \sf\ Reference \: of \: occuring \: of \: outcomes \: is \\ \sf\ shown \: in \: table:$

$\boxed{\begin{tabular}{ c || c} \bf{Observation} & \bf{Times of occurring}\\ \cline{1-2}\sf{39} & \sf{1} \\ \cline{1-2}\sf{40} & \sf{2}\\ \cline{1-2}\sf{42} & \sf{1}\\ \cline{1-2}\sf{46} & \sf{1}\\ \cline{1-2}\sf{48} & \sf{1}\\ \cline{1-2}\sf{52} & \sf{3}\\ \cline{1-2}\sf{54} & \sf{1}\\ \cline{1-2}\sf{60} & \sf{1}\\ \cline{1-2}\sf{62} & \sf{1}\\ \cline{1-2}\sf{86} & \sf{1}\\ \cline{1-2}\sf{98} & \sf{1}\end{tabular}}$

$\normalsize\sf\ From \: the \: observations \: it \: is \: quite \: simple \: \\ \sf\ to \: know \: that \: 52(3)\: has \: highest \: times \: of \: occurring \\\\\\\maltese\:\:\boxed{\sf\ Mode = Maximum \: number \: of \: occuring}\\\\\\ \normalsize\ : \implies\sf\ Mode = 52 \\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{ Mode = 52}}}}$

Answer 2

Answer

$\Large\frak{\underline{\underline{Correct \: Question:}}}$

In a maths test given to 15 students the following marks out of 100 are recorded:

41,39,48,52,46,62,54,40,96,52,98,40,42,52,60

Find the Mean,Median and Mode of data.

$\rule{170}2$

$\Large\frak{\underline{\underline{Your~Answer:}}}$

$\underline{\bigstar\:\textsf{Mean \: of \: data:}}$

$\maltese\:\:\:\boxed{\normalsize\sf\ Mean = \frac{Sum \: of \: all \: observations}{Number \: of \: total \: observations}}\\\\\\\normalsize\ : \implies\sf\ Mean = \scriptsize\frac{(41 +39+48+52+46+62+54+40+96+52+98+40+42+52+60)}{15}\\\\\\\normalsize\ : \implies\sf\ Mean = \frac{\cancel{822}}{\cancel{15}}\\\\\\\normalsize\ : \implies\sf\ Mean = 54.8\\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{ Mean = 54.8}}}}$

$\underline{\bigstar\:\textsf{Median \: of \: data:}}\\ \\ \normalsize\sf\bullet\: Arrange \: the \: data \: into \: ascending \: order:\\ 39,40,40,42,46,48,52,52,52,54,60,62,96,98\\ \sf\bullet\ Number \: of \: observations(n) = 15 = odd \: number \\\\\maltese\:\:\boxed{\sf\ Median = \frac{n +1}{2}^{th} \: term}\\\\\\\normalsize\ : \implies\sf\ Median = \frac{15 + 1}{2}^{th} \: term\\\\\\\normalsize\ : \implies\sf\ Median = \frac{\cancel{16}}{\cancel{2}}^{th} \: term\\\\\\\normalsize\ : \implies\sf\ Median = 8^{th} \: term\\\\\\\normalsize\ : \implies\sf\ Median = 52\\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{Median = 52 }}}}$

$\underline{\bigstar\:\textsf{Mode \: of \: data:}}\\\\ \sf\ Reference \: of \: occuring \: of \: outcomes \: is \\ \sf\ shown \: in \: table:$

$\boxed{\begin{tabular}{ c || c} \bf{Observation} & \bf{Times of occurring}\\ \cline{1-2}\sf{39} & \sf{1} \\ \cline{1-2}\sf{40} & \sf{2}\\ \cline{1-2}\sf{42} & \sf{1}\\ \cline{1-2}\sf{46} & \sf{1}\\ \cline{1-2}\sf{48} & \sf{1}\\ \cline{1-2}\sf{52} & \sf{3}\\ \cline{1-2}\sf{54} & \sf{1}\\ \cline{1-2}\sf{60} & \sf{1}\\ \cline{1-2}\sf{62} & \sf{1}\\ \cline{1-2}\sf{86} & \sf{1}\\ \cline{1-2}\sf{98} & \sf{1}\end{tabular}}$

$\normalsize\sf\ From \: the \: observations \: it \: is \: quite \: simple \: \\ \sf\ to \: know \: that \: 52(3)\: has \: highest \: times \: of \: occurring \\\\\\\maltese\:\:\boxed{\sf\ Mode = Maximum \: number \: of \: occuring}\\\\\\ \normalsize\ : \implies\sf\ Mode = 52 \\\\\\\normalsize\ : \implies{\underline{\boxed{\sf \red{ Mode = 52}}}}$