Question

find the missing frequency f , if the mode of the given data is 154
class : 120-130 130-140 140-150 150-160 160-170 170-180
frequency : 2 8 12 f 8 7

Answer 1

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

Given data is:

$\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ 120 - 130 & 2  \\130 - 140 & 8 \\140 - 150 & 12 \\150 - 160 & f \\160 - 170 & 8\\170 - 180 & 7 \end{array}\end{gathered}$

We know,

Formula of Mode is

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

Where,

$\: \: \: \: \: \: \: \: \sf \bull \: \: \: l \: is \: lower \: limit \: of \: modal \: class$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \: f_0 \: is \: frequency\: preceeding\: the \: modal \: class$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \: f_1 \: is \: frequency\: of\: the \: modal \: class$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \: f_2 \: is \: frequency\: succeeding\: the \: modal \: class$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \: h \: is \: height\: of\: the \: modal \: class$

Here,

Given that

• Mode = 154

Hence,

• Modal class = 150 - 160

Thus,

$\: \: \: \: \: \: \: \: \sf \bull \: \: \:l = 150$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \:f_0 = 12$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \:f_1 = f$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \:f_2 = 8$

$\: \: \: \: \: \: \: \: \sf \bull \: \: \:h = 10$

On Substituting all these values in formula,

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

$\rm :\longmapsto\:{{\bf{154 = 150 + \bigg(\dfrac{f - 12}{2f - 12 - 8} \bigg) \times 10 }}}$

$\rm :\longmapsto\:{{\bf{154 - 150 = \bigg(\dfrac{f - 12}{2f - 20} \bigg) \times 10 }}}$

$\rm :\longmapsto\:{{\bf{4 = \bigg(\dfrac{f - 12}{f - 10} \bigg) \times 5 }}}$

$\rm :\longmapsto\:4f - 40 = 5f - 60$

$\rm :\longmapsto\:4f - 5f = 40 - 60$

$\bf\implies \:f = 20$

Additional Information :-

$\dashrightarrow\sf Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}$

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i}$

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$

Answer 2

Answer:

Step-by-step explanation:

First we write down the known values:

Given

Mode=154

f1=?

f2=?

f0=?

l=?

h=?

Write down the table

Class      Frequency

120-130       2

130-140       8

140-150       12

150-160        f

160-170        8

170-180        7

From this we can find the values hence

f1= f

(Why?: because in the question they are saying that ''The mode of the distribution is 154" so the highest value in this should be 'f' Because it lies in between 150-160)

f2=8

f0=12

l=150

h=10

x=l+(f1-fo)/(2f1-fo-f2)xh

154=150+ (f-12) / (2f -12-8) x 10

4= 120- 2f / 2f -20

4(2f-20)=120- 2f

8f-80=120- 2f

2f=40

f=20