Statistics Test - 46

To find the median class:

First add the total number of frequencies.

Here, $$f = 50$$

$$= \dfrac{total \quad number \quad of \quad frequency}{2}$$

 $$=\dfrac { (50 )}{2} = 25$$

Hence, the value 25 lies in the class interval 31-41 from the cummulative frequency table.

Here frequency of class interval $$10-12$$ is maximum.

So, the modal class is $$10-12$$

Now $$ l $$ = lower limit of modal class $$= 10$$

$$f_1 =$$ frequency of modal class $$= 6$$

$$f_0 =$$ frequency of class preceding the modal class $$= 3$$

$$f_2 =$$ frequency of class succeeding the modal class $$= 3$$

$$h =$$ class width $$= 2$$

So, Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$

                 $$=10+\left(\displaystyle\frac{6-3}{2\times 6-3-3}\right)\times 2$$

                 $$=10+\left(\displaystyle\frac{3}{12-6}\right)\times 2$$

                 $$=10+\left(\displaystyle\frac{3}{6}\right)\times 2$$

                 $$=10+1$$

                 $$=11$$

Hence, mode of the given frequency distribution is $$11$$

Here frequency of class interval $$75-85$$ is maximum.

So, it is the modal class

Now $$ l $$ = lower limit of modal class $$= 75$$

$$f_1 =$$ frequency of modal class$$ = 10$$

$$f_0 = $$frequency of class preceding the modal class $$= 3$$

$$f_2 =$$ frequency of class succeeding the modal class $$= 7$$

$$h =$$ class interval width $$= 10$$

So, Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$

$$=75+\left(\displaystyle\frac{10-3}{2\times 10-3-7}\right)\times 10$$

$$=75+\left(\displaystyle\frac{7}{20-10}\right)\times 10$$

$$=75+7$$

$$=82$$

Mode$$=82$$

Here the frequency of class interval $$70-80$$ is the maximum.

So, it is the modal class

Now 

$$ l $$ = lower limit of modal class $$= 70$$

$$f_1 =$$ frequency of modal class $$= 24$$

$$f_0 =$$ frequency of class preceding the modal class $$= 10$$

$$f_2 =$$ frequency of class succeeding the modal class $$= 5$$

$$h = $$class interval width $$= 10$$

Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$

           $$=70+\left(\displaystyle\frac{24-10}{2\times 24-10-5}\right)\times 10$$

           $$=70+\left(\displaystyle\frac{14}{48-15}\right)\times 10$$

           $$=70+4.24$$

           $$=74.24$$

Mode $$\approx 74$$

Here frequency of class interval $$60-70$$ is maximum.

So, it is the modal class

Now,$$ l = $$  lower limit of modal class

$$ = 60$$

$$f_1 =$$ frequency of modal class 

$$= 40$$

$$f_0 =$$ frequency of class preceding the

modal class 

$$= 10$$

$$f_2 =$$ frequency of class succeeding the

modal class 

$$= 30$$

$$h =$$ class interval width

$$ = 10$$

So, Mode$$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$

$$=60+\left(\displaystyle\frac{40-10}{2\times 40-10-30}\right)\times 10$$

$$=60+\left(\displaystyle\frac{30}{80-40}\right)\times 10$$

$$=60+7.5$$

$$=67.5$$

Mode$$=67.5$$

Over	Run
1-5	0
5-9	2
9-13	4
13-17	10
17-21	3
21-25	11
25-29	0

$$Mode=l+\left( \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ { 2f }_{ 1 }-{ f }_{ 0- }{ f }_{ 2 } }  \right) \times h\\ l=lower\quad limit\quad of\quad modal\quad class\\ h=size\quad of\quad class\quad interval\\ { f }_{ 1 }=frequency\quad of\quad the\quad modal\quad class\\ { f }_{ 0 }=frequency\quad of\quad the\quad class\quad preceeding\quad the\quad modal\quad class\\ { f }_{ 2 }=frequency\quad of\quad the\quad class\quad succeeding\quad the\quad modal\quad class\\ Maximum\quad Frequency=11\\ Modal\quad class=21-25\\ l=21,{ f }_{ 1 }=11\quad { f }_{ 0 }=3\quad { f }_{ 2 }=0,h=2\\ \\ Mode\quad =21+\left( \cfrac { 11-3 }{ 22-3-0 }  \right) \times 2\\ =21+\left( \cfrac { 8 }{ 19 }  \right) \times 2\\ =21+\cfrac { 16 }{ 19 } \\ =21+0.84\\ =21.84Ans$$

Marks Students $$50-60$$ $$1$$ $$60-70$$ $$2(f_0)$$ $$70-80$$ $$x(f_1)$$ $$80-90$$ $$4(f_2)$$

$$ Mode=l+\left( \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ { 2f }_{ 1 }-{ f }_{ 0}-{ f }_{ 2 } }  \right) \times h\\ \quad \implies 76=70+\left( \cfrac { x-2 }{ 2x-6 }  \right) \times 10\\ \implies \cfrac { 6}{ 10 } =\cfrac { x-2 }{ 2x-6 } \\6x-18=5x-10\\ \therefore x=8$$