Question

The median of the following data is 525. Find the values of x and y l, if the total frequency is 100.

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Answer 1

$\boxed{\begin{array}{c|c|c} \bf \underline{ \: \: Class \: intervals \: \: }&\bf \underline{ \: \: Frequency \: \: }&\bf \underline{ \: \: Cumulative \: frequency \: \: }\\ \: \: \: \: \: 0 - 100&2&2\\100 - 200&5&7\\200 - 300& \rm x& \rm 7 \pmb{+} x\\300 - 400&12&19 \pmb + \rm x\\400 - 500&17&36 \pmb + \rm x\\500 - 600&20&56 \pmb + \rm{x }\\600 - 700& \rm y&56 \pmb + \rm{x \pmb+ y}\\700 - 800&9&65 \pmb + \rm{x \pmb+ y}\\800 - 900& \rm 7&72 \pmb + \rm{x \pmb+ y}\\ \: \: \: 900 - 1000& \rm 4&76\pmb + \rm{x \pmb+ y}\end{array}}$

Given:-

$\rm{n = 100 }$

$\rm{76 \pmb+ x \pmb+ y = 100}$

$\rm{x \pmb + y = 100 \pmb- 76}$

$\rm{x \pmb + y = 24 \xrightarrow{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }(1)}$

Median is 525,

Which lies in the Class intervals 500 – 600

$l = 500$

$f = 20$

$\rm cf = 36 + x$

$h = 100$

Formula:-

$: \longmapsto \underline{ \boxed{ \mathrm{Median} = l \pmb+ \left \lgroup\begin{array}{cc} \\ \dfrac{ \dfrac{n}{2} - \mathrm{cf}}{f} \\ \\ \end{array}\right \rgroup h}}$

$: \longmapsto \: 525 = 500 + \bigg \{ \dfrac{50 - 36 - \mathrm{x}}{20} \bigg \} \pmb \times 100$

$: \longmapsto \: 525 = 500 + \bigg \{ \dfrac{14- \mathrm{x}}{ \cancel{20}} \bigg \} \pmb \times \cancel{100}$

$: \longmapsto \: 525 - 500 = \big(14 - \mathrm{ x} \big) \times 5$

$: \longmapsto \: 5 \mathrm{x}= 70 - 25$

$: \longmapsto \: 5 \mathrm{x}= 45$

$: \longmapsto \: \mathrm{x}= \cancel{\dfrac{45}{5}}$

$: \longmapsto \: \mathrm{x}= 9$

[From 1]

$: \longmapsto \: 9 + \mathrm{y}= 24$

$: \longmapsto \: \mathrm{y}= 24 - 9$

$: \longmapsto \: \mathrm{y}= 15$