Question

SPAMMERS WILL BE REPORTED

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

FULL EXPLANATION NEEDED ​

Answer 1

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

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

$\huge\underline\mathrm{SOLUTION:-}$

Mean

• Mean of data = Sum of all observation/Number of observation

$\implies \mathsf {\frac{41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 50}{15} }$

➠ $\mathsf {\frac{822}{15} }$

➠ 54.8

Median

Arranging data in ascending order,

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

Number of observations = n = 15 (Odd Number)

Median = (n + 1/2)^th Observation

➠ (15 + 1/2)^th Observation

➠ 8th Observation

➠ 52

Mode

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

${\underline{\boxed{\sf{Here:-}}}}$

• 39 Occurs 1 time
• 40 Occurs 2 times
• 41 Occurs 1 time
• 42 Occurs 1 time
• 46 Occurs 1 time
• 48 Occurs 1 time
• 52 Occurs 3 times
• 54 Occurs 1 time
• 60 Occurs 1 time
• 62 Occurs 1 time
• 96 Occurs 1 time
• 98 Occurs 1 time

Since, 52 Occurs maximum number of times.

ThereFore:

• Mode = 52.

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