CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 16 - MCQExams.com

X	10-20	20-30	30-40	40-50	50-60	60-70
Y	2	3	6	5	5	5

Class interval	mid values (x)	f	d=x-35	fd
10-20	15	2	-20	-40
20-30	25	3	-10	-30
30-40	35	6	0	0
40-50	45	5	10	50
50-60	55	2	20	40
60-70	65	2	30	60
Total		22		80

$$\overline{x} =A+\dfrac{\sum fd}{N}=35+\dfrac{(80)}{32}=38.6$$  

Median is the middle most value of a series. So when the series has odd number of elements then median can be calculated easily but when the series has even number of elements then the series has two middle values, so median is calculated either by taking out the average of both the value or the median is the (N+1)/2 th element of the series.

List of pulse rates: $$64,70,74,80,92$$

From the given graph, median corresponds to $$300$$ count.

$$\begin{matrix} \Rightarrow \frac { { dy } }{ { dx } } =\frac { { { y^{ 2 } }-2y } }{ { x-1 } }  \\ \Rightarrow \frac { { dy } }{ { { y^{ 2 } }-2y } } =\frac { { dx } }{ { \left( { x-1 } \right)  } }  \\ \Rightarrow \int { \frac { { dy } }{ { y\left( { y-2 } \right)  } } = } \frac { { dx } }{ { x-1 } }  \\ \Rightarrow \int { \frac { { dy } }{ { y\left( { y-2 } \right)  } } = } \log  \left( { x-1 } \right) +c \\ Let \\ \Rightarrow \frac { 1 }{ { y\left( { y-2 } \right)  } } =\frac { A }{ y } +\frac { B }{ { \left( { y-2 } \right)  } }  \\ \Rightarrow 1=Ay-2A+By \\ \Rightarrow 1+2A \\ \Rightarrow A=\frac { { -1 } }{ 2 }  \\ and\, \, \left( { A+B } \right) y=0, \\ A+B=0\to (i) \\ Put\, \, A=\frac { { -1 } }{ 2 } \, \, in\, \, equation(i) \\ \frac { { -1 } }{ 2 } +B=0,\, \, B=\frac { 1 }{ 2 }  \\ \Rightarrow \int { \frac { A }{ y } dy+\int { \frac { B }{ { \left( { y-2 } \right)  } } dy=\log  \left( { x-1 } \right) +c }  }  \\ \Rightarrow \frac { { -1 } }{ 2 } \int { \frac { { dy } }{ y } +\frac { 1 }{ 2 } \int { \frac { 1 }{ { \left( { y-2 } \right)  } } =\log  (x-1 } +c }  \\ \Rightarrow \frac { { -1 } }{ 2 } \log  y+\frac { 1 }{ 2 } \log  \left( { y-2 } \right) =\log  \left( { x-2 } \right) +\log  c \\ \Rightarrow \frac { { -1 } }{ 2 } \left[ { \log  \left( { \frac { { y-2 } }{ y }  } \right)  } \right] =\log  \left( { x-1 } \right) c \\ \Rightarrow \log  \sqrt { \frac { { y-2 } }{ y }  } =\log  \left( { x-1 } \right) c \\ \therefore \sqrt { \frac { { \left( { y-2 } \right)  } }{ y }  } =\left( { x-1 } \right) c \\ \Rightarrow \frac { y }{ y } \, \, \frac { { -2 } }{ y } ={ \left( { x-1 } \right) ^{ 2 } }{ c^{ 2 } } \\ \Rightarrow 1-{ \left( { x-1 } \right) ^{ 2 } }{ c^{ 2 } }=\frac { 2 }{ y }  \\ \therefore y=\frac { 2 }{ { -1{ { \left( { x-1 } \right)  }^{ 2 } }{ c^{ 2 } } } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Ans. \\  \end{matrix}$$