Question

50 points!!!
Someone help pls, I can’t understand it and it’s due tomorrow :c

Answer 1

${\large{\textsf{\textbf{\underline{\underline{Question \: 1 :}}}}}}$

$\star\:{\underline{\underline{\sf{\purple{Solution:}}}}}$

❍ Arrange the given data in order either in ascending order or descending order.

2, 3, 4, 7, 9, 11

❍ Number of terms in data [n] = 6 which is even.

As we know,

$\star \: \sf Median_{(when \: n \: is \: even)} = {\underline{\boxed{\sf{\purple{ \dfrac{ { \bigg (\dfrac{n}{2} \bigg)}^{th}term +{ \bigg( \dfrac{n}{2} + 1 \bigg)}^{th} term } {2} }}}}}$

$\sf Median_{(when \: n \: is \: even)} ={ \dfrac{ { \bigg (\dfrac{6}{2} \bigg)}^{th}term +{ \bigg( \dfrac{6}{2} + 1 \bigg)}^{th} term } {2} }$

$\sf Median_{(when \: n \: is \: even)} ={ \dfrac{ {3}^{rd} term +{ \bigg( \dfrac{6 + 2}{2} \bigg)}^{th} term } {2} }$

$\sf Median_{(when \: n \: is \: even)} ={ \dfrac{ {3}^{rd} term +{ \bigg( \cancel{ \dfrac{8}{2}} \bigg)}^{th} term } {2} }$

$\sf Median_{(when \: n \: is \: even)} ={ \dfrac{ {3}^{rd} term +{ 4}^{th} term } {2} }$

• Putting,

3rd term as 4 and the 4th term as 7.

$\longrightarrow \: \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ 4 + 7 } {2} }$

$\longrightarrow \: \sf Median_{(when \: n \: is \: even)} ={ \dfrac{ 11} {2} }$

$\longrightarrow \: \sf Median_{(when \: n \: is \: even)} = \purple{5.5}$

${\large{\textsf{\textbf{\underline{\underline{Question \: 2 :}}}}}}$

$\star\:{\underline{\underline{\sf{\red{Solution:}}}}}$

❍ Arrange the given data in order either in ascending order or descending order.

1, 2, 3, 4, 5, 6, 7

❍ Number of terms in data [n] = 7 which is odd.

As we know,

$\star \: \sf Median_{(when \: n \: is \: odd)} = {\underline{\boxed{\sf{\red{ { \bigg( \frac{n + 1}{2} \bigg)}^{th} term}}}}}$

$\sf Median_{(when \: n \: is \: odd)} = {{ \bigg(\dfrac{ 7 + 1 } {2} \bigg) }}^{th} term$

$\sf Median_{(when \: n \: is \: odd)} = { \bigg(\cancel{\dfrac{8}{2}} \bigg)}^{th} term$

$\sf Median_{(when \: n \: is \: odd)} ={ 4}^{th} term$

• Putting,

4th term as 4.

$\longrightarrow \: \sf Median_{(when \: n \: is \: odd)} = \red{ 4}$

${\large{\textsf{\textbf{\underline{\underline{Question \: 3 :}}}}}}$

$\star\:{\underline{\underline{\sf{\green{Solution:}}}}}$

The frequency distribution table for calculations of mean :

$\begin{gathered}\begin{array}{|c|c|c|c|c|c|c|} \hline \rm x_{i} &\rm 3&\rm 1&\rm 7&\rm 4&\rm 6&\rm 2 \rm \\ \hline\rm f_{i} &\rm 4&\rm 6&\rm 2&\rm 2 & \rm 1&\rm 1 \\ \hline \rm f_{i}x_{i} &\rm 12&\rm 6&\rm 14&\rm 8&\rm 6&\rm \rm 2 \\ \hline \end{array} \\ \end{gathered}$

☆ Calculating the $\sum f_{i}$

$\implies 4 + 6 + 2 + 2 + 1 + 1$

$\implies 16$

☆ Calculating the $\sum f_{i}x_{i}$

$\implies 12 + 6 + 14 + 8 + 6 + 2$

$\implies 48$

As we know,

Mean by direct method :

$\: \: \boxed{\green{{ { \overline{x} \: = \sf \dfrac{ \sum \: f_{i}x_{i}}{ \sum \: f_{i}}}}}}$

here,

• $\sum f_{i}$ = 16

• $\sum f_{i}x_{i}$ = 48

By putting the values we get,

$\sf \longrightarrow \overline{x} \: = \: \dfrac{48}{16}$

$\sf \longrightarrow \overline{x} \: = \green{3}$

${\large{\textsf{\textbf{\underline{\underline{Note\: :}}}}}}$

• Swipe to see the full answer.

$\begin{gathered} {\underline{\rule{290pt}{3pt}}} \end{gathered}$