Question

2. The mean of the following data is 42. Find the missing frequencies
and y if the sum of frequencies is 100.
Class interval
0-10 10-20 20-30 30-40 40-50 50-6000-70 70-80
Frequency
7
10
x
13
y
10
14
9​

Answer 1

$\sf\large\underline{Given:-}$

$\sf{\implies The\:mean=42}$

$\sf{\implies Sum\:_{(frequency)}=100}$

$\sf\large\underline{To\:Find:-}$

$\sf{\implies The\: missing\: frequency=?}$

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

To calculate missing frequency at first we have to make a table by making table we have to find sum of three currency and some of of FX then by applying formula to to calculate the missing frequency by setting up equations:-

$\sf\green{\implies Refer\:to\: attachment\: here:-}$

$\sf\small\underline\red{Calculation\:for\:1st\: equation:-}$

$\sf\large\underline{Sum\:_{(frequency)}=100:-}$

$\tt{\implies x+y+63=100}$

$\tt{\implies x+y=100-63}$

$\tt{\implies x+y=37-----(i)}$

$\sf\small\underline\red{Calculation\:for\:2nd\: equation:-}$

$\tt{\implies Mean=\dfrac{\sum\:f\:x}{\sum\:f}}$

$\tt{\implies 42=\dfrac{2775+25x+45y}{x+y+63}}$

$\tt{\implies 42x+42y+2646=2775+25x+45y}$

$\tt{\implies 42x-25x+42y-45y=2775-2646}$

$\tt{\implies 17x-3y=129---(ii)}$

In eq (i) multiplying by 17 then subract:-]

$\tt{\implies 17x+17y=629}$

$\tt{\implies 17x-3y=129}$

By solving we get here,

$\tt{\implies 20y=500}$

$\tt{\implies y=25}$

Putting the value of y=25 in eq (i):]

$\tt{\implies x+y=37}$

$\tt{\implies x+25=37}$

$\tt{\implies x=37-25}$

$\tt{\implies x=12}$

$\sf\large{Hence'}$

$\sf{\implies The\: missing\: frequency=25\:and\:12}$

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• The mean of the following data is 42 .

• The sum of frequencies are 100 .

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

• The missing frequencies,

1. x
2. y

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

$\begin{tabular}{| c | c | c | c |} Class interval & Frequency (f_i) & x_i & f_i x_i \\ \cline{1 - 4} 0 - 10 & 7 & 5 & 35 \\ 10 - 20 & 10 & 15 & 150 \\ 20 - 30 & x & 25 & 25x \\ 30 - 40 & 13 & 35 & 455 \\ 40 - 50 & y & 45 & 45y \\ 50 - 60 & 10 & 55 & 550 \\ 60 - 70 & 14 & 65 & 910 \\ 70 - 80 & 9 & 75 & 675 \\ \cline{1 - 4} & \Sigma{f_i} = 63 + x + y & & \Sigma{f_i x_i} = 2775 + 25x + 45y \\ \cline{1 - 4} \end{tabular}$

✞︎ According to the question,

$\huge\red\checkmark$ N ($\bf\red{\Sigma{f_i}}$) = 100

⇢ 63 + x + y = 100

⇢ x + y = 100 - 63

⇢ x + y = 37 ------(1)

☯︎ We know that,

$\purple\bigstar\:\bf{\red{\overbrace{\underbrace{\green{Mean\:=\:\dfrac{\Sigma{f_i\:x_i}}{\Sigma{f_i}}\:}}}}}$

$\rm{{\orange{:\implies}}\:42\:=\:\:\dfrac{2775\:+\:25x\:+\:45y}{100}\:}$

$\rm{{\green{:\implies}}\:4200\:=\:2775\:+\:25x\:45y\:}$

$\rm{{\orange{:\implies}}\:4200\:-\:2775\:=\:25x\:+\:45y\:}$

$\rm{{\green{:\implies}}\:1425\:=\:5\:(5x\:+\:9y)\:}$

$\rm{{\orange{:\implies}}\:\dfrac{1425}{5}\:=\:5x\:+\:9y\:}$

$\bf{{\green{:\implies}}\:5x\:+\:9y\:=\:285\:}$ -----(2)

➪ Multiple equation (1) into 9, we get

• 9x + 9y = 333 -------(a)

➪ Now substracting equation (2) from equation (a),

⇒ 9x + 9y - 5x - 9y = 333 - 285

⇒ 4x = 48

⇒ x = 48/4

⇒ x = 12

☞︎︎︎ Putting the value of 'x' in equation (1), we get

➟ 12 + y = 37

➟ y = 37 - 12

➟ y = 25

__________________________

$\huge\red\therefore$ The missing frequency x is "12 " .

$\huge\red\therefore$ The missing frequency y is "25 " .