Question

plz solve this problem, I'll foll.ow you if I find your answer he.lpful.​

Answer 1

$\huge\tt\underline\red{Ans}\underline {w}\underline\purple{er}\underline{:=}\orange {113.33}$

$\huge \underbrace \mathfrak \red{\bigstar Explanation}$

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\begin{gathered}\begin{array}{|c|c|c|}\cline{1-3}Marks\;Obtained & Number\;of\;Students(f) & C.F \\ \\ 3 140-150 & 12 & 12 \\ \\ {3} 130-140 & 27-12=15 & 27 \\ \\ -3120-130 & 60-27=33 & 60 \\ \\ {3} 110-120 & 105-60=45 & 105 \\ \\ {3} 100-110 & 124-105=19 & 124 \\ \\ {3} 90-100 & 141-124=17 & 141 \\ \\ {3} 80-90 & 150-141=9 & 150 \end{array}\end{gathered}$

N = 150

$\tt{\rightarrow\dfrac{n}{2}=75}$

$\because\blue{\boxed{Median\;class\;is\;110-120}}$

Here we have :-

$\tt{\rightarrow\dfrac{n}{2}=75}$

l = 110, f = 45 , C.F = 60 and h = 10

$\tt{\rightarrow Median=\dfrac{n/2-C.F}{f}\times h}$

$\tt{\rightarrow 110 + \dfrac{75-60}{45}\times 10}$

$\tt{\rightarrow 110 + \dfrac{150}{45}\times 10}$

= 110 + 3.33

= 113.33

${\boxed{\green{\:Median\:=113.33\: marks}}}$