Question

Twenty students of Class XI have secured the following marks: 11, 12, 14, 11, 16, 11, 17, 16, 17, 14
17, 18, 20, 14, 20, 17, 20, 17, 14, 20.
Present the data (i) As a Individual Series
(ii) As a frequency array.
(iii) Continuous series taking interval of 10- 15

Answer 1

Answer:Question

Explanation:

Answer 2

Answer:

Answer:

$\begin{tabular}{|c|c|}\cline{1-2}\sf Class Interval&\sf Frequency \\\cline{1-2}\sf0-10&\sf5\\\cline{1-2}\sf10-20&\sf8\\\cline{1-2}\sf20-30&\sf15\\\cline{1-2}\sf30-40&\sf20\\\cline{1-2}\sf40-50&\sf14\\\cline{1-2}\sf50-60&\sf8\\\cline{1-2}\sf60-70&\sf3\\\cline{1-2}\end{tabular}$

⠀

Here maximum frequency is 20, the class corresponding to this 30 - 40

⠀

So modal class is 30-40

Now lower limit of modal class ( l ) = 30

Frequency ( f₁ ) of modal class = 15

Frequency ( f₀ ) of preceding modal class = 20

Frequency ( f₂ ) of the class succeeding modal class = 14

Class size ( h ) = 10

⠀

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Mode= l+\bigg[\dfrac{f_o-f_1}{2f_o-f_1-f_2}\bigg]\times h\\\\\\:\implies\sf Mode=30+\bigg[\dfrac{20-15}{2(20)-15-14}\bigg]\times10\\\\\\:\implies\sf Mode= 30+\bigg[\dfrac{5}{40-29}\bigg]\times10\\\\\\:\implies\sf Mode=30+\bigg[\dfrac{10\times5}{11}\bigg]\\\\\\:\implies\sf Mode= 30+\dfrac{50}{11}\\\\\\:\implies\sf Mode= 30+4.54\\\\\\:\implies\underline{\boxed{\sf\ Mode= 34.54}}$

⠀

$\therefore\:\underline{\textsf{Hence, Mode of the given data is \textbf{34.54}}}$