Question

Please answer it
➡️ Wrong Answer will be reported​

Answer 1

S O L U T I O N :

We have class & frequency :

Firstly, make a data of frequency distribution.

$\begin{tabular}{|c|c|} \cline{1-2} \multicolumn{2} {|c|} {\bf DATA} \\ \cline{1-2} \multicolumn{1} {|c|} { \bf Class} & \multicolumn {1} {|c|} {\bf Frequency} \\ \cline{1-2} \sf 0-5 &\sf 5 \\ \cline{1-2} \sf 5-10 & \sf 12 \\ \cline{1-2} \sf 10-15 & \sf 18 \\ \cline{1-2} \sf 15-20 & \boxed{\bf{28} = f_1} \\ \cline{1-2} \sf 20-25 & \sf 17 \\ \cline{1-2} \sf 25 -30 & \sf 12 \\ \cline{1-2} \sf 30-35 & \sf 8 \\ \cline{1-2} \end{tabular}$

$\underline{\bf{Explanation\::}}$

As we know that formula of the mode a/q;

$\boxed{\bf{Mode = l + \bigg(\frac{f_1 - f_0}{2f_1 - f_0 -f_2} \bigg) \times h}}$

Where as,

• (l), Lower limit of the model - class = 15
• (f1), Frequency of the model - class = 28
• (f0), This frequency of the class before the model - class = 18
• (f2), This frequency of the class after the model - class = 17
• (h), Size of the model class = 5

Now,

Putting the value on given formula :

$\mapsto\tt{Mode = 15 + \bigg(\dfrac{28 - 18 }{2(28) - 18 - 17} \bigg) \times 5}$

$\mapsto\tt{Mode = 15 + \bigg(\dfrac{28 - 18 }{56 - 35} \bigg) \times 5}$

$\mapsto\tt{Mode = 15 + \bigg(\dfrac{10 }{21} \bigg) \times 5}$

$\mapsto\tt{Mode = 15 + \cancel{\dfrac{50 }{21}}}$

$\mapsto\tt{Mode = 15 + 2.38}$

$\mapsto\bf{Mode = 17.38}$

Thus,

The mode of the following frequency distribution will be 17.38 .

Answer 2

The formula of mode

$\sf{\boxed{\color{gold}{l + ( \frac{f1 - f0}{2f1 - f0 - f2} ) \times h}}}$

Here,

$\sf{l=15 , f _{1} = 28 ,\: f _{0} = 18, \: f _{2} = 17 ,\: h = 5}$

Now, putting the values in the formula of mode, we get,

$\sf{\rightarrow \: 15 + ( \frac{28 - 18}{2 \times 28 - 18 - 17} \times 5}$

$\sf{\rightarrow \:15 + \frac{10}{56 - 18 - 17} \times 5}$

$\sf{\rightarrow \: 15 + \frac{10}{21} \times 5 }$

$\sf{\rightarrow 15 + \cancel \frac{50}{21} }$

$\sf{\rightarrow 15 + 2.38 }$

$\sf{\rightarrow 17.38}$

Hence, the mode of the data is 17.38.