Question

If L=50,fm=16,f2=14,f1=8,h=10 then find Mode *


​

Answer 1

Given:

Lower limit of modal class, l = 50,frequency of modal class, $f_{m}$ = 16,

Succeeding frequency of modal class, $f_{2}$ = 14,  preceding frequency of modal class, $f_{1}$ = 8 and width if the modal class, h = 10

We have to find, the mode of the given data.

Solution:

We know that:

Mode = $l+\dfrac{f_{m} -f_{1}}{2f_{m} -f_{1}-f_{2}} \times h$

          = $50+\dfrac{16 -8}{2(16) -8-14} \times 10$

         = $50+\dfrac{8}{32 -22} \times 10$

         = 50 + 8

         = 58

∴ Mode = 58

Thus, the mode of the given data is 58.

Answer 2

Given:

L=50, fm=16, f2=14, f1=8, h=10

To find:

Find mode

Solution:

From given, we have the data as follows.

L = 50, fm = 16, f2 = 14, f1 = 8, h = 10

There is a direct formula using which this problem can be solved.

So, the formula that relates the mode with the given parameters is given as follows.

$Mode=l+\dfrac{f_m-f_1}{2f_m-f_1-f_2} \times h$

where l = limit of modal class = 50

$f_m$ = frequency of modal class = 16

$f_2$ = succeeding frequency of modal class = 14

$f_1$ = preceding frequency of modal class = 8

h = width of the modal class = 10

Substitute the given values in the above equation.  

$Mode=50+\dfrac{16-8}{2\times 16-8-14} \times 10\\Mode=50+\dfrac{8}{10} \times 10\\Mode=50+8\\Mode=58$

Therefore, the value of the mode is 58.