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The media of the following data is 525. Find the value of x and y, if the total frequency is 100.

Class interval Frequency (fᵢ)
0 - 100 2
100 - 200 5
200 - 300 x
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 y
700 - 800 9
800 - 900 7
900 - 100 4
​

The Media Of The Following Data Is 525 Find The Value Of X And Y If The Total Frequency Is 100Class Interval Frequency Fᵢ0 100 2100 200 5200 300 X300 400 12400 class=

Sagot :

SalmonWorldTopperidn

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-}}[/tex]

The frequency distribution table is as follow

[tex]\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}[/tex]

Given that,

  • Sum of all frequencies = 100

So,

[tex]\rm :\longmapsto\: \sum \: f \: = \: 100[/tex]

[tex]\rm :\longmapsto\: 76 + x + y \: = \: 100[/tex]

[tex]\rm :\longmapsto\: x + y \: = \: 100 - 76[/tex]

[tex]\rm\implies \:\boxed{ \: \tt{ x + y = 24 \: }} - - - (1)[/tex]

Further given that,

  • Median of the series, M = 525

We know, Median is evaluated by using the formula,

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

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,

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

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

[tex]\dashrightarrow\sf 525 - 500 = 5(50 - 36 - x)[/tex]

[tex]\dashrightarrow\sf 25 = 5(14 - x)[/tex]

[tex]\dashrightarrow\sf 5 =14 - x[/tex]

[tex]\bf\implies \:x = 9[/tex]

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

[tex]\rm :\longmapsto\:9 + y = 24[/tex]

[tex]\bf\implies \:y = 15[/tex]

Hence,

[tex]\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{x = 9} \\ \\ &\bf{y = 15} \end{cases}\end{gathered}\end{gathered}[/tex]

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