Question

Please solve the attachment!!​

Answer 1

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

The frequency distribution table is as follow

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: cumulative \: frequency\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 100&\sf 2&\sf2\\\\\sf 100 - 200 &\sf 5&\sf7\\\\\sf 200-300 &\sf x&\sf7 + x\\\\\sf 300-400 &\sf 12&\sf19 + x\\\\\sf 400-500 &\sf 17&\sf36 + x\\\\\sf 500-600 &\sf 20&\sf56 + x\\\\\sf 600-700 &\sf y&\sf56 + x + y\\\\\sf 700 - 800&\sf 9&\sf65 + x + y\\\\\sf 800-900&\sf 7&\sf72 + x + y\\\\\sf 900-1000&\sf 4&\sf76 + x + y\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Given that,

Sum of all frequencies = 100

So,

$\rm :\longmapsto\: \sum \: f \: = \: 100$

$\rm :\longmapsto\: 76 + x + y \: = \: 100$

$\rm :\longmapsto\: x + y \: = \: 100 - 76$

$\rm\implies \:\boxed{ \: \tt{ x + y = 24 \: }} - - - (1)$

Further given that,

• Median of the series, M = 525

We know, Median is evaluated by using the formula,

$\rm :\longmapsto\:\boxed{ \sf Median, \: M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$

Here,

• l denotes lower limit of median class

• h denotes width of median class

• f denotes frequency of median class

• cf denotes cumulative frequency of the class preceding the median class

• N denotes sum of frequency

According to the given distribution table,

We have

• Median class is 500 - 600

So, we have

• l = 500,

• h = 100,

• f = 20,

• cf = 36 + x

• N = 100

By substituting all the given values in the formula,

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

$\dashrightarrow\sf 525= 500 + \Bigg \{100 \times \dfrac{ \bigg( \dfrac{100}{2} - (36 + x) \bigg)}{20} \Bigg \}$

$\dashrightarrow\sf 525 - 500 = 5(50 - 36 - x)$

$\dashrightarrow\sf 25 = 5(14 - x)$

$\dashrightarrow\sf 5 =14 - x$

$\bf\implies \:x = 9$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:9 + y = 24$

$\bf\implies \:y = 15$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{x = 9} \\ \\ &\bf{y = 15} \end{cases}\end{gathered}\end{gathered}$

Answer 2

Answer:

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

The frequency distribution table is as follow

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: cumulative \: frequency\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 100&\sf 2&\sf2\\\\\sf 100 - 200 &\sf 5&\sf7\\\\\sf 200-300 &\sf x&\sf7 + x\\\\\sf 300-400 &\sf 12&\sf19 + x\\\\\sf 400-500 &\sf 17&\sf36 + x\\\\\sf 500-600 &\sf 20&\sf56 + x\\\\\sf 600-700 &\sf y&\sf56 + x + y\\\\\sf 700 - 800&\sf 9&\sf65 + x + y\\\\\sf 800-900&\sf 7&\sf72 + x + y\\\\\sf 900-1000&\sf 4&\sf76 + x + y\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Given that,

Sum of all frequencies = 100

So,

$\rm :\longmapsto\: \sum \: f \: = \: 100$

$\rm :\longmapsto\: 76 + x + y \: = \: 100$

$\rm :\longmapsto\: x + y \: = \: 100 - 76$

$\rm\implies \:\boxed{ \: \tt{ x + y = 24 \: }} - - - (1)$

Further given that,

Median of the series, M = 525

We know, Median is evaluated by using the formula,

$\rm :\longmapsto\:\boxed{ \sf Median, \: M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$

Here,

l denotes lower limit of median class

h denotes width of median class

f denotes frequency of median class

cf denotes cumulative frequency of the class preceding the median class

N denotes sum of frequency

According to the given distribution table,

We have

Median class is 500 - 600

So, we have

l = 500,

h = 100,

f = 20,

cf = 36 + x

N = 100

By substituting all the given values in the formula,

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

$\dashrightarrow\sf 525= 500 + \Bigg \{100 \times \dfrac{ \bigg( \dfrac{100}{2} - (36 + x) \bigg)}{20} \Bigg \}$

$\dashrightarrow\sf 525 - 500 = 5(50 - 36 - x)$

$\dashrightarrow\sf 25 = 5(14 - x)$

$\dashrightarrow\sf 5 =14 - x$

$\bf\implies \:x = 9$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:9 + y = 24$

$\bf\implies \:y = 15$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{x = 9} \\ \\ &\bf{y = 15} \end{cases}\end{gathered}\end{gathered}$