Math, asked by Anonymous, 8 months ago

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Please answer the question given in the attachment with full explanation...
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Answered by AdorableMe
213

Q 6. Solution :-

\begin{tabular}{|c|c|}\cline{1-2}Class Interval&Frequency\\\cline{1-2} 0-6&4\\\cline{1-2}6-12&x\\\cline{1-2}12-18&5\\\cline{1-2}18-24&y\\\cline{1-2}24-30&1\\\cline{1-2}\end{tabular}

Calculating the C.F :-

\begin{tabular}{|c|c|c|}\cline{1-3}Class interval&Frequency&C.F.\\\cline{1-3} 0-6&4&4\\\cline{1-3}6-12&x&4+x\\\cline{1-3}12-18&5&9+x\\\cline{1-3}18-24&y&9+x+y\\\cline{1-3}24-30&1&10+x+y\\\cline{1-3}&\sum f=20& \\\cline{1-3}\end{tabular}

◘ 10 + x + y = 20

⇒  x + y = 20 - 10

⇒  x + y = 10          . . . (i)

◘  Median = 14.4  

So, Median class is 12 - 18

Now,

  • Lower limit of the Median Class, l = 12
  • Class width, h = 6
  • Cumulative frequency of the class before the Median Class, cf = 4 + x
  • n/2 = 20/2 = 10
  • Frequency of the Median Class, f = 5

We know,

\displaystyle{\sf{Median = l+\frac{\frac{n}{2}-cf }{f}\times h }}

\displaystyle{\sf{\implies Median =12+\frac{10-(4+x)}{5}  \times6}}\\\\\displaystyle{\sf{\implies 14.4 =12+ \frac{6-x}{5} \times 6 }}\\\\\displaystyle{\sf{\implies 14.4 = 12+\frac{36-6x}{5} }}\\\\\displaystyle{\sf{\implies 14.4 = \frac{60+36-6x}{5} }}\\\\\displaystyle{\sf{\implies 72 =96-6x }}\\\\\displaystyle{\sf{\implies 12=16-x }}\\\\\boxed{\boxed{\sf{\implies  x=4}}}

Putting the value of x in eq. (i) :-

\sf{x + y = 10}\\\\\sf{ \implies 4 + y = 10}\\\\\sf{\implies y = 10 - 4}\\\\\boxed{\boxed{\sf{\implies y = 6}}}


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Answered by Anonymous
246

Given :

  • The median of the distribution is 14.4

  • Total frequency is 20.

To Find :

  • Find the value of x and y.

Solution:

\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency&\sf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0-6&\sf 4&\sf 4 \\\\\sf 6-12 &\sf x &\sf 4 + x \\\\\sf 12-18&\sf 5&\sf 9 + x\\\\\sf 18-24 &\sf y &\sf 9 + x + y\\\\\sf 24-30 &\sf 1 &\sf 10 + x + y\end{array}}

Median = 14.4....(given)

So, Median class = 12-18

  • Lower limit of median class (l) = 12

  • Class interval (h) = 6

  • CF = 4 + x

  • Frequency of the median class = 5

  • n/2 = 20/2 = 10

\large{\boxed{\green{\bf Median =\ l\ +\ \dfrac{\frac{n}{2}\ -\ cf}{f}\ \times\ h}}}

 \sf 14.4 =\ 12\ +\ \dfrac{10 - (x + 4)}{5}\ \times\ 6

 \sf 14.4  =\ 12 \ + \dfrac{6 \times (10 - x - 4)}{5}

\sf 12 = 36 - 6x

\sf 6x = 36 - 12

\sf 6x = 24

\pink{\sf x = 4}

Now,

➨10 + x + y = 20

➨x + y = 10

➨y = 10 - 4.....(putting x = 4)

\pink{\sf y = 6}

Therefore, value of x and y are 4 and 6 respectively.


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