Question

4. Find the quartiles for given data set a) 6 874599 b) 6 8745893 5. Compute the z-score if a) x 40, x 30,s 5

Answer 1

Solution : ( 4 )

( a )

$\mathrm{First\:Quartile\:\left(lower\:quartile\right)\::}$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$4,\:5,\:6,\:7,\:8,\:9,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}4&5&6&7&8&9&9\\ 1&2&3&4&5&6&7\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}7$

$\mathrm{Since\:the\:number\:of\:terms\:is\:odd,\:take\:the\:elements\:below\:the\:middle\:one,\:that\:is,\:the\:lower\:}3\mathrm{\:elements}$$4,\:5,\:6$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$4,\:5,\:6$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}4&5&6\\ 1&2&3\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}3$

$\mathrm{Since\:the\:number\:of\:terms\:is\:odd,\:the\:median\:is\:the\:middle\:element\:of\:the\:sorted\:set.}$

$5$

$\mathrm{The\:median\:of\:the\:ascending\:set}:\quad 5$

$\mathrm{Median\:of\:}4,\:5,\:6:\quad 5$

$\mathrm{Hence,\quad\:}\mathrm{First\:Quartile\:\left(lower\:quartile\right)\:of\:\:\:}6,\:8,\:7,\:4,\:5,\:9,\:9:\quad \bold{5}$

$\mathrm{and}$

$\mathrm{Second\:Quartile\:\:(median):}$

$\mathrm{Arrange\:the\:data\:in\:ascending\:order\::\quad\:}4,\:5,\:6,\:7,\:8,\:9,\:9$

$\mathrm{Here,\:\:median=7}$

$\mathrm{Hence,\:\:Second\:Quartile\:=\bold{7}}$

$\mathrm{and}$

$\mathrm{Third\:Quartile\:\left(upper\:quartile\right)}\::$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$4,\:5,\:6,\:7,\:8,\:9,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}4&5&6&7&8&9&9\\ 1&2&3&4&5&6&7\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}7$

$\mathrm{Since\:the\:number\:of\:terms\:is\:odd,\:take\:the\:elements\:above\:the\:middle\:one,\:that\:is,\:the\:upper\:}3\mathrm{\:elements}.$$8,\:9,\:9$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$8,\:9,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}8&9&9\\ 1&2&3\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}3$

$\mathrm{Since\:the\:number\:of\:terms\:is\:odd,\:the\:median\:is\:the\:middle\:element\:of\:the\:sorted\:set.}$

$9$

$\mathrm{The\:median\:of\:the\:ascending\:set}:\quad 9$

$\mathrm{Median\:of\:}8,\:9,\:9:\quad 9$

$\mathrm{Hence,\quad\:}\mathrm{Third\:Quartile\:\left(upper\:quartile\right)\:of\:\:\:}6,\:8,\:7,\:4,\:5,\:9,\:9:\quad \bold{9}$

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( b )

$\mathrm{First\:Quartile\:\left(lower\:quartile\right)\::}$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$3,\:4,\:5,\:6,\:7,\:8,\:8,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}3&4&5&6&7&8&8&9\\ 1&2&3&4&5&6&7&8\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}8$

$\mathrm{Take\:the\:lower\:}4\mathrm{\:terms}$

$3,\:4,\:5,\:6$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$3,\:4,\:5,\:6$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}3&4&5&6\\ 1&2&3&4\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}4$

$\mathrm{Since\:the\:number\:of\:terms\:is\:even,\:the\:median\:is\:the\:average\:of\:the\:two\:middle\:elements:\quad }4,\:5$

$\mathrm{Arithmetic\:Mean\:\left(average\right)\:of\:}4,\:5=\frac{4+5}{2}=\frac{9}{2}=4.5$

$\mathrm{Hence,\quad\:}\mathrm{First\:Quartile\:\left(lower\:quartile\right)\:of\:}6,\:8,\:7,\:4,\:5,\:8,\:9,\:3:\quad \bold{4.5}$

$\mathrm{and}$

$\mathrm{Second\:Quartile\:\:(median):}$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$3,\:4,\:5,\:6,\:7,\:8,\:8,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}3&4&5&6&7&8&8&9\\ 1&2&3&4&5&6&7&8\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}8$

$\mathrm{Since\:the\:number\:of\:terms\:is\:even,\:the\:median\:is\:the\:average\:of\:the\:two\:middle\:elements:\quad }6,\:7$

$\mathrm{Arithmetic\:Mean\:\left(average\right)\:of\:}6,\:7=\frac{6+7}{2}=\frac{13}{2}=6.5$

$\mathrm{Hence,\quad\:}\mathrm{Second\:Quartile\:\left(median\right)\:of\:}6,\:8,\:7,\:4,\:5,\:8,\:9,\:3:\quad \bold{6.5}$

$\mathrm{and}$

$\mathrm{Third\:Quartile\:\left(upper\:quartile\right)}\::$

$\mathrm{Arrange\:the\:terms\:in\:ascending\:order}$

$3,\:4,\:5,\:6,\:7,\:8,\:8,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}3&4&5&6&7&8&8&9\\ 1&2&3&4&5&6&7&8\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}8$

$\mathrm{Take\:the\:upper\:}4\mathrm{\:terms}$

$7,\:8,\:8,\:9$

$\mathrm{Count\:the\:number\:of\:terms\:in\:the\:data\:set}$

$\begin{Bmatrix}7&8&8&9\\ 1&2&3&4\end{Bmatrix}$

$\mathrm{The\:number\:of\:terms\:in\:the\:data\:set\:is\:\:}4$

$\mathrm{Since\:the\:number\:of\:terms\:is\:even,\:the\:median\:is\:the\:average\:of\:the\:two\:middle\:elements:\quad }8,\:8$

$\mathrm{Arithmetic\:Mean\:\left(average\right)\:of\:}8,\:8=\frac{8+8}{2}=\frac{16}{2}=8$

$\mathrm{The\:median\:of\:the\:ascending\:set}:\quad 8$

$\mathrm{Median\:of\:}7,\:8,\:8,\:9:\quad 8$

$\mathrm{Hence,\quad\:}\mathrm{Third\:Quartile\:\left(upper\:quartile\right)\:of\:}6,\:8,\:7,\:4,\:5,\:8,\:9,\:3:\quad \bold{8}$

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Solution : ( 5 )

( a )

$\mathrm{z-score}=\frac{x-\bar{x}}{s}=\frac{40-30}{5}=\frac{10}{5}=2$

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( b )

$\mathrm{z-score}=\frac{x-\mu }{\sigma }=\frac{50-60}{4}=\frac{-10}{5}=-2.5$