Question

If the mean of the following frequency distribution is 65.6. Find the missing frequencies f1 and f2 by using assumed mean method.

ᴄʟᴀss ғʀᴇǫᴜᴇɴᴄʏ
10-30 5
30-50 8
50-70 f1
70-90 20
90-110 f2
110-130 2
ᴛᴏᴛᴀʟ 50​

Answer 1

Answer

Here we are provided with mean of the given frequency is 65.6

we know,

$\sf 5 + 8 + f_1 + 20 + f_2 + 2 = 50$

$\sf 35 + f_1+ f_2 + = 50$

$\sf f_2 =15 - f_1 \: \: \: \: ...(1)$

Firstly we have to prepare a Frequency distribution table,

$\begin{gathered} \boxed{\begin{array}{|c|c|c|c|c|}\cline{1-5}\bf Class&\bf Frequency (f_i)&\bf class \: mark (x_i) & \bf d_i = (x_i - A)& \bf d_ix_i\\\cline{1-5}\sf 10-39&\sf 5&\sf20 & \sf -40 & \sf -200\\\cline{1-5}\sf 30-50&\sf 8&\sf 40& \sf -20 & \sf -160\\\cline{1-5}\sf 50-70&\sf f_1 &\sf 60 & \sf 0 & \sf 0\\\cline{1-5}\sf 70-90&\sf 20& \sf 80 & \sf 20 & \sf 400\\\cline{1-5}\sf 90 - 110&\sf 15 - f_1&\sf 100 & \sf 40 & \sf 600 - 40f_1\\\cline{1-5}\sf 110-130&\sf 2& \sf 120 & \sf 60 & \sf 120 \\\cline{1-5}\sf &\sf \sum f_i= 50& & & \sf \sum f_id_i = 760 - 40f_1\\\cline{1-5}\end{array}}\end{gathered}$

By using Assume mean method that is,

$\boxed{ \sf \overline{x} = A + \dfrac{\sum f_id_i}{\sum f_i}}$

According to the question,

A = 60

Σf$\sf _{i}$x$\sf _{i}$ = 760 - 40f$\sf _{i}$

Σf$\sf _{i}$ = 50

mean = 65.6

By substituting all the given values in the formula,

$\dashrightarrow \: \: \sf \overline{x} = A + \dfrac{\sum f_id_i}{\sum f_i}$

$\dashrightarrow \: \: \sf 65.6 = 60 + \dfrac{760 - 40f_1}{50}$

$\dashrightarrow \: \: \sf 65.6 - 60 = \dfrac{760 - 40f_1}{50}$

$\dashrightarrow \: \: \sf 5.6 \times 50 = 760 - 40f_1$

$\dashrightarrow \: \: \sf 280 = 760 - 40f_1$

$\dashrightarrow \: \: \sf 760 - 280 = 40f_1$

$\dashrightarrow \: \: \sf 480 = 40f_1$

$\dashrightarrow \: \: \sf f_1 = \dfrac{480}{40}$

$\dashrightarrow \: \: \boxed{\sf f_1= 12}$

now, substituting f₁ = 12 in eq (1),

$\dashrightarrow \: \: \sf f_{2}=15 - f_1$

$\dashrightarrow \: \: \sf f_{2}=15 - 12$

$\dashrightarrow \: \: \boxed{ \sf f_{2}=3}$

Thus, f₁ = 12 and f₂ = 3

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

It is given that,

$:  \implies \tt \: \red{ \bf \: \sum f_i= 50}$

$:  \implies \tt \: 5 + 8 + f_1 + 20 + f_2 + 2 = 50$

$:  \implies \tt \: 35 + f_1 + f_2 = 50$

$:  \implies \tt \: \boxed{ \purple{ \bf \: f_1 + f_2 = 15}} - - (i)$

Now,

Prepare a frequency distribution table as follow

$\begin{gathered}\begin{gathered} \boxed{\begin{array}{|c|c|c|c|c|}\bf Class&\bf Frequency (f_i)&\bf class \: mark (x_i) & \bf d_i = (x_i - A)& \bf f_id_i\\\sf 10-30&\sf 5&\sf20 & \sf -40 & \sf -200 \\ \sf 30-50&\sf 8&\sf 40& \sf -20 & \sf -160\\\sf 50-70&\sf f_1 &\sf 60 & \sf 0 & \sf 0\\\sf 70-90&\sf 20& \sf 80 & \sf 20 & \sf 400\\\sf 90 - 110&\sf f_2&\sf 100 & \sf 40 & \sf 40f_2\\\sf 110-130&\sf 2& \sf 120 & \sf 60 & \sf 120 \\\sf &\sf \sum f_i= 50& & & \sf \sum f_id_i = 160 + 40f_2\\\end{array}}\end{gathered} \end{gathered}$

Now, we have

$:  \implies \tt \: \red{\sum f_i= 50}$

$:  \implies \tt \: \blue{A = 60}$

$:  \implies \tt \: \purple{\sum f_id_i= 160 + 40f_2}$

$:  \implies \tt \: \green{ \overline{x} \: = 65.6}$

Now,

We know that,

$:  \implies \tt \: \boxed{ \red{ \bf \overline{x} = A + \dfrac{\sum f_id_i}{\sum f_i}}}$

$:  \implies \tt \: 65 .6 \: = 60 + \dfrac{160 + 40f_2}{50}$

$:  \implies \tt \: 5.6 = \dfrac{160 + 40f_2}{50}$

$:  \implies \tt \: 280 = 160 + 40f_2$

$:  \implies \tt \: 40f_2 = 120$

$:  \implies \boxed{ \red{\tt \: f_2 = 3}}$

On substituting the value in equation (i), we get

$:  \implies \tt \: 3 + f_1 \: = 15$

$:  \implies \tt \: f_1 = 15 - 3$

$:  \implies \boxed{ \purple{ \bf \: f_1 = 12}}$