Question

calculate median, mode, median class interval 1-4,4-7,7-10,10-13,13-16,16-19 frequency 6,30,40,16,4,4​

Answer 1

Step-by-step explanation:

Mode is

mode=l+{(f1-f0)/2f1-f0-f2}h

mode=7+ (10/34)3

mode=7+30/34

mode=7+15/17

mode=7+0.88

mode=7.88

Median

median=l+ {(n/2-cf)/f}h

median=7+{(50-36)/40}3

median=7+(14×3)/40

median=7+42/40

median=7+21/20

median=7+1.05

median=8.05

Median class interval is 3

Answer 2

Given: Frequency distribution table

To find: Median , mode , median class

Step-by-step explanation:

Class interval            f                      CF

1-4                               6                      6

4-7                             30                   36

7-10                            40                   76

10-13                           16                   92

13-16                            4                    96

16-19                            4                   100

Now to find median as we know that

$median =L + \dfrac{\dfrac{n}{2} -CF}{f} \times h$

where L is the lower limit , h is the class  interval , f is the frequency of median class CF is the preceding cumulative frequency and n is the total frequencies

We have  $\dfrac{n}{2} =50$

Therefore C.F.= 36 , h = 3 , L = 7 and f = 40

So we have

$median= 7+\dfrac{50-36}{40} \times 3\\\\\Rightarrow median= 7+\dfrac{14}{40} \times 3 = 7+1.05 = 8.05$

as median lies between the class interval 7 - 10 therefore it is its median class

now we have to find the mode

As we know

$mode= L+ \dfrac{f_1-f_0}{2f_1-f_0-f_2} \times h$

where $f_1$ is the highest frequency of modal class,  $f_0$ is the frequency of preceding modal class, $f_2$ is the frequency of succeeding modal class and L is the Lower limit of modal class

Therefore

 $f_1$ = 40 ,  $f_0$ = 30 , $f_2$ = 16 , L=7 ,h= 3

So

$mode= 7+\dfrac{40-30}{2\times 40-30-16} \times 3\\\\\Rightarrow mode= 7+\dfrac{10}{34} \times 3= 7+ 0.88 = 7.88$

Hence , Median is 8.05 ,Mode is 7.88 and median class is 7-10