Question

The mean of the following frequency table 50. But the frequencies f; and f 2 in class 20-40 and 60-80 are missing. Find the missing frequencies. Class: 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Total Frequency: 17 fi 32 fi 19 120 (CBSE 2006C)​

Answer 1

Question:

• The mean of the following frequency table 50. But the frequencies f 1 and f2 in class 20-40 and 60-80 are missing. Find the missing frequencies.

Given:

• The mean of frequency 50 .but the frequencies f1 and f2 in class 20 -40 and 60 - 80 are missing.

To find:

• Total frequency.

Solution:

As it is given that total of frequency is 120.

$\sf \: So, \: \: N=\: ∑ f {\footnotesize{1} }\: = \: 120 \: \: \: \: \: \: \: \: (given)$

$\dashrightarrow \sf \: 68 \: + \: f {\footnotesize{1}} \: + {\: f \footnotesize{2}} \: = \: 120$

$\dashrightarrow \sf \: f{ \footnotesize{1} }\: + \: f {\footnotesize{2} }\: = \: 120 \: - \: 68$

$\dashrightarrow \sf \: f{ \footnotesize{1} }\: + \: f {\footnotesize{2}} \: = 52$

$\sf Or \: \: \: \: f {\footnotesize{2}} \: = \: 52 \: - {f \footnotesize{1}}$

$\sf \: Mean \: = \: \dfrac{ \:∑{f \footnotesize{i}} \:{ X{\footnotesize{i}}}}{ \:∑f \footnotesize {i} }$

$\sf \: Mean \: = \: \dfrac{3480 \: + \: 30 f {\footnotesize{1}} \: + \: 70 \:{ f \footnotesize{2}}}{120}$

$\sf \: 50 \: = \: \dfrac{3480 \: + \: 30 f {\footnotesize{1}} \: + \: 70 \:{ f \footnotesize{2}}}{120}$

$\rm \: [ Given \: that \: mean = \: 50 ]$

$\sf \: 50 \times120= 3480 \: + \: 30f{ \footnotesize{1} }+70(52 \: - \: f \footnotesize{1})$

$\sf[ \: Put \: f{ \footnotesize2} \: = \: 52 \: - \: f{ \footnotesize{1}} \: ]$

$\sf \: 6000 = 3480 \: + \: 30f {\footnotesize{1}}\: + \: 3640 \: - 70f \footnotesize{1}$

$\sf \: 6000 \: = \: 7120 \: - \: 40f \footnotesize1$

$\sf \: 6000 \: - \: 7120 = - 40f \footnotesize1$

$\sf\bf - \: 1120 = - 40f {\footnotesize{1}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1120 = 40f \footnotesize{1}$

$\rm[Negative \: sign \: from \: both \: sides \: cancel]$

$\sf \: f{ \footnotesize{1}} = \dfrac {\cancel{1120} \: \: ^{28} }{ \cancel{40}}$

$\sf \large\: f {\footnotesize{1}} \: = {\boxed{ \boxed{ \pink{28}}}}$

$\rm It \: has \: been \: found \: that \: f {\footnotesize{1}} = 28$

$\rule{300px}{ \: 5e}$

Also, it was earlier established that

$\sf \large\: f{ \footnotesize{1}} \: + \: f {\footnotesize{2}} = 52$

$\dashrightarrow\sf\: 28 \: + \: f {\footnotesize{2} }= 52 \: \: \: \: \: \: \: \: \: \: \: \: \: [ \: f{\footnotesize{1} } = 28 \: ]$

$\dashrightarrow \sf\:f {\footnotesize{2} }= 52 \: - \: 28$

$\sf \large \dashrightarrow {\boxed{ \boxed{ \pink{24}}}}$

$\rm\large \: So, \: f {\footnotesize{1}} = 28 \: \: and \: \: f {\footnotesize{2}} = 24.$

Your answer is 28 and 24.

Answer 2

ans : F1=28

F2=24

thank you