Question

0-2 frequency-1
2-4frequency-2
4-6frequency-1
6-8frequency-5
8-10frequency-6
10-12frequency-2
12-14frequency-3
so Q-what is the mode? ​

Answer 1

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

Given data is

$\begin{gathered} \begin{array}{|c|c|} \bf{Class \: Interval} & \bf{Frequency} \\ \\ 0 - 2 & 1  \\2 - 4 & 2 \\4 - 6 & 1 \\6 - 8 & 5 \\8 - 10 & 6\\ 10 - 12 & 2\\ 12 - 14 & 3 \end{array}\end{gathered}$

We know,

Mode of data is given by

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

where,

l is the lower limit of Modal class

h is the height of the interval

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

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

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

Now, from given data we have

Modal class is 8 - 10

So,

$\rm \: l = 8$

$\rm \: h = 2$

$\rm \: {f_0} = 5$

$\rm \: {f_1} = 6$

$\rm \: {f_2} = 2$

So, on substituting the values, we get

$\rm \: Mode = 8 + \dfrac{6 - 5}{12 - 5 - 2} \times 2$

$\rm \: Mode = 8 + \dfrac{1}{12 - 7} \times 2$

$\rm \: Mode = 8 + \dfrac{1}{5} \times 2$

$\rm \: Mode = 8 + 0.4$

$\rm\implies \:\rm \: Mode = 8.4$

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

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

Mode:-

A mode is that value among the observations which occurs most often, that is, the value of the observations having the maximum frequency.

$\boxed{ \begin{array}{ c|c } \textsf{ \textbf{Class Interval }}&\textsf{ \textbf{ Frequency}} \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }& \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ 0 - 2&1 \\ 2 - 4&2 \\ 4 - 6&1 \\ 6 - 8& \: \: \: 5_{f_{0}}\\ 8 - 10& \: \: \: 6_{f_{1}} \\ 10 - 12& \: \: \: 2_{f_{2}} \\ 12 - 14&3 \end{array}}$

$\quad \star \underbrace{ \underline{ \boxed{ \mathrm{Mode} = l + \left\lbrace \dfrac{{f_{1}}-{f_{0}}}{{2f_{1}}-{f_{0}-{f_{2}}}}\right \rbrace \times h} }}$

We know,

• $l$ = lower limit of the modal class,
• $h$ = size of the class Interval,
• ${f_{1}}$ = frequency of the modal class,
• ${f_{0}}$ = frequency of the class preceding the modal class,
• ${f_{2}}$ = frequency of the class succeeding the modal class.

Let's substitute the given values,

$l = 8$

$f _{1} = 6$

$h = 2$

$f _{0} = 5$

$f _{2} = 2$

$: \longmapsto 8 + \left\lbrace \dfrac{6 - 5}{2(6)-{5 - {2}}}\right \rbrace \times 2$

$: \longmapsto 8 + \left\lbrace \dfrac{1}{12-{5 - {2}}}\right \rbrace \times 2$

$: \longmapsto 8 + \left\lbrace \dfrac{1}{12{- 7 {}}}\right \rbrace \times 2$

$: \longmapsto 8 + \dfrac{1}{ 5} \times 2$

$: \longmapsto 8 + \dfrac{2}{ 5}$

$: \longmapsto 8.4$

∴The Mode of the data above is 8.4