Question

7. Find the mode of the following frequency distribution: Class 0 - 20 20-40 40-60 60-80 80 - 100 100-120 Frequency 8. 9 26 23 20 14 and he moda follow framan​

Answer 1

$\large\underline{\sf{Solution-}}$

Given data is

$\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ \\ 0 - 20 & 8  \\  \\20 - 40 & 9  \\ \\40 - 60 & 26 \\ \\ 60 - 80 & 23  \\ \\80 - 100 & 20\\ \\ 100 - 120 & 14 \end{array}\end{gathered}$

Now, from above data we concluded that

$\rm \: f_0 = 9$

$\rm \: f_1 = 26$

$\rm \: f_2 = 23$

$\rm \: h = 23$

$\: l = 40$

We know,

$\boxed{ \boxed{\rm{ \: \: \: Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h \: \: \: }}}$

where,

$\: l \: \rm \: is \: lower \: limit \: of \: modal \: class$

$\: h \: \rm \: is \: height \: of \: modal \: class$

$\rm \: {f_0} \:  is \: frequency \: of \: class \: preceding \: modal \: class$

$\rm{f_1} \:  is \: frequency \: of \: modal \: class$

$\rm{f_2} \: is \: frequency \: of \: class \: succeeding \: modal \: class$

So, on substituting the values, we get

$\rm \: Mode = 40 + \bigg(\dfrac{26 - 9}{2 \times 26 - 9 - 23} \bigg) \times 20$

$\rm \: Mode = 40 + \bigg(\dfrac{17}{52 - 32} \bigg) \times 20$

$\rm \: Mode = 40 + \bigg(\dfrac{17}{20} \bigg) \times 20$

$\rm \: Mode = 40 +17$

$\rm\implies \:\boxed{ \rm{ \:\bf \: Mode = 57 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

1. Mean Using Direct Method :-

$\boxed{ \rm{ \: \: Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: \: }}$

2. Mean Using Short Cut Method :-

$\boxed{ \rm{ \: \: Mean =A \: + \: \dfrac{ \sum f_i d_i}{ \sum f_i} \: \: }}$

3. Mean Using Step Deviation Method :-

$\boxed{ \rm{ \: \: Mean =A \: + \: \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: \: }}$

Answer 2

Answer:

Step-by-step explanation: