Question

solve this please .fast as possible​

Answer 1

Solution :

$\begin{tabular}{|c |c | c|}\cline{1-3}\bf \underline {Class interval}&\bf\underline {Frequency (f)}&\bf\underline {C.F}\\\cline{1-3}\sf 100-200 & \sf 11 & \sf 11 \\\sf 200-300 & \sf 12 & \sf p \\\sf 300-400 &\sf 10 & \sf 33\\\sf 400-500 &\sf q &\sf 46\\\sf 500-600 & \sf 20 &\sf 66 \\\sf 600-700 &\sf 14 &\sf 80 \\\cline{1-3}\end{tabular}$

p = 11 + 12 = 23

and 33 + q = 46

➨ 46 - 33

➨ 13

Hence, the value of p and q are 23 and 13 respectively.

Here,

• Maximum C.F = 20

• Class corresponding to this frequency = 500-600

Hence, the modal class is 500-600.

and n = 80

So, n/2 = 80/2 = 40

• The CF is just greater than 40 is 46.

• The corresponding class is 400-500.

Therefore, the median class is 400-500.

Answer 2

Answer:

$\begin{tabular}{|c|c|c|}\cline{1-3}Class Interval& Frequency & C.f. \\\cline{1-3}100 - 200 & 11 & 11 \\\cline{1-3}200 - 300 & 12 & p\\\cline{1-3}300 - 400 & 10 & 33\\\cline{1-3}400 - 500 & q & 46\\\cline{1-3}500 - 600 & 20 & 66\\\cline{1-3}600 - 700 & 14 & 18 \\\cline{1-3}\end{tabular}$

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

Here,

p = 11 + 12 = 23

And

33 + q = 46

q = 46 - 33 = 13

$\therefore\:\underline{\textsf{So, the value of p is 23 and q is 13.}}$

$\rule{130}{1}$

Now,

Here the maximum class frequency is 20 and the class corresponding to this frequency is 500 - 600.

$\therefore\:\underline{\textsf{So, the\:modal\:class\:is \textbf{500 - 600. }}}$

$:\implies\sf N=\sum\limits f\\\\\\:\implies\sf \dfrac{N}{2} = \dfrac{\sum\limits f}{2}\\\\\\:\implies\sf \dfrac{N}{2} = \dfrac{80}{2}\\\\\\:\implies\sf{N} = 40$

⠀

The cf is just greater than 40 is 46 and the corresponding class is 400 - 500.

$\therefore\:\underline{\textsf{Hence, Median class is \textbf{400 - 500}}}.$