Question

tell the answer but with solution pic I will mark as brainliest​

Answer 1

$\colorbox{gold} {\text{ \bf♕ Brainliest answer }}$

$\rule{300pt}{0.1pt}$

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\begin{array}{|c|c|c|} \hline \\ \rm x & \rm Frequency (f) & \rm c f \\ \\ \hline 15 & 3 & 3 \\ \hline 17 & 5 & 8 \\ \hline 20 & 9 & 17 \\ \hline 22 & 4 & 21 \\ \hline 25 & 6 & 27 \\ \hline 30 & 10 & 37 \\ \hline \end{array}$

No. of observations, N=37, which is odd.

$\text{So, median \( \rm =\left(\dfrac{N+1}{2}\right)^{t h} \)Observation.}$

$\text{ \( \rm\ =\left(\dfrac{37+1}{2}\right)^{t h} \) Observation}$

$\text{ \( \rm\ =\left(\dfrac{38}{2}\right)^{t h} \) Observation}$

$\text{ \( =19^{\text {th }} \) Observation}$

$\rm \therefore 19^{th}$ student as per the cumulative frequency table has 22 marks. Hence, 22 is the median score.

Answer 2

Step-by-step explanation:

$\tiny\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

$\rule{300pt}{0.1pt}$

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\begin{gathered}\begin{array}{|c|c|c|} \hline \\ \rm x & \rm Frequency (f) & \rm c f \\ \\ \hline 15 & 3 & 3 \\ \hline 17 & 5 & 8 \\ \hline 20 & 9 & 17 \\ \hline 22 & 4 & 21 \\ \hline 25 & 6 & 27 \\ \hline 30 & 10 & 37 \\ \hline \end{array}\end{gathered}$

No. of observations, N=37, which is odd.

$\text{So, median \( \rm =\left(\dfrac{N+1}{2}\right)^{t h} \)Observation.}$

$\text{ \( \rm\ =\left(\dfrac{37+1}{2}\right)^{t h} \) Observation}$

$\text{ \( \rm\ =\left(\dfrac{38}{2}\right)^{t h} \) Observation}$

$\text{ \( =19^{\text {th }} \) Observation}$

$\rm \therefore 19^{th}$ student as per the cumulative frequency table has 22 marks. Hence, 22 is the median score.