Question

Compute the median for following data :-

More than 150 - 0

More than 140 - 12

More than 130 - 27

More than 120 - 60

More than 110 - 105

More than 100 - 124

More than 90 - 141

More than 80 - 150

Answer 1

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

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

N = 150

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

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

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

${\boxed{Here\;Median\;=113.33\; marks}}$

Answer 2

Question :--- Find the Median of Following data ?

Formula used :---

Median = L + [{(N/2) -cf} / F ] * C

where ,

L = lower boundary of the median class.

N = sum of frequencies.

cf = cumulative frequency before the median class.

F = frequency of the median class.

C = The size of the median class.

______________________________

❁❁ Refer To Image First .. ❁❁

Solution :---

→ N = Sum of Frequency = (12 + 15 + 33 + 45 + 19 + 17 + 9) = 150 ..

Now, Lets find Median class of Grouped data .

→ Median class = (N/2)

→ Median class = 150/2 = 75..

So, we have to look class interval that cover 75 marks. ( that will be our Median class. ).

→ Looking at the Cumulative Frequency , we can see that 75 comes in 110-120..

So, out Median class is 110-120...

______________________________

Now,

→ L = lower boundary of the median class = 110

N = sum of frequencies = 150

cf = cumulative frequency before the median class = 60.

F = frequency of the median class = 45

C = The size of the median class = 10.

Putting all values now in Formula we get,

→ Median = L + [{(N/2) -cf} / F ] * C

→ Median = 110 + [( 75 - 60) / 45 ] * 10

→ Median = 110 + [ (15/45) ]* 10

→ Median = 110 + 10/3

→ Median = 110 + 3 + 1/3

→ Median = 113 + /13

→ Median = 113(1/3)..

Hence, Median of Grouped given data is 113(1/3)..