Question

the mode of the follow frequency distribution is 55. find the value of x and y.

Answer 1

Answer:

y = 5 and x = f - 43 (where f is sum of all frequencies)

Step-by-step explanation:

Given: Mode = 55

Data is given in Grouped Frequency Distribution table.

This mean give mode is of continuous series.

Formula of Mode of Continuous series is given by,

$Mode=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

where, l = lower limit of modal class

           h = Width/length of modal class

           $f_1$ = frequency of modal class

           $f_0$ = frequency of class preceding modal class

           $f_2$ = frequency of class succeeding  modal class

Here, Modal class is 45 - 60 (because mode lies in this class interval)

⇒ l = 45 , h = 15 , $f_1$ = 15 , $f_0$ = y , $f_2$ = 10

substituting these values in formula, we get

$55=45+\frac{15-y}{2\times15-y-10}\times 15$

$55-45=\frac{15-y}{20-y}\times 15$

$10=\frac{15-y}{20-y}\times 15$

$\frac{10}{15}=\frac{15-y}{20-y}$

$\frac{15-y}{20-y}=\frac{2}{3}$

$3\times(15-y)=2\times(20-y)$

$45-3y=40-2y$

$-3y+2y=40-45$

$-y=-5$

$y=5$

To find Value of x we should know sum of all frequencies.

Let say sum of all frequencies be f

so, steps to find value of x

Sum of all frequencies = f

6 +7 + y + 15 + 10 + x = f

6 + 7 + 5 + 15 +10 + x = f

x + 43 = f

x = f - 43

Therefore, Value of x is (sum of frequencies) - 43.  

Answer 2

Step-by-step explanation:

Answer:

y = 5 and x = f - 43 (where f is sum of all frequencies)

Step-by-step explanation:

Given: Mode = 55

Data is given in Grouped Frequency Distribution table.

This mean give mode is of continuous series.

Formula of Mode of Continuous series is given by,

Mode=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times hMode=l+

2f

1

−f

0

−f

2

f

1

−f

0

×h

where, l = lower limit of modal class

h = Width/length of modal class

f_1f

1

= frequency of modal class

f_0f

0

= frequency of class preceding modal class

f_2f

2

= frequency of class succeeding modal class

Here, Modal class is 45 - 60 (because mode lies in this class interval)

⇒ l = 45 , h = 15 , f_1f

1

= 15 , f_0f

0

= y , f_2f

2

= 10

substituting these values in formula, we get

55=45+\frac{15-y}{2\times15-y-10}\times 1555=45+

2×15−y−10

15−y

×15

55-45=\frac{15-y}{20-y}\times 1555−45=

20−y

15−y

×15

10=\frac{15-y}{20-y}\times 1510=

20−y

15−y

×15

\frac{10}{15}=\frac{15-y}{20-y}

15

10

=

20−y

15−y

\frac{15-y}{20-y}=\frac{2}{3}

20−y

15−y

=

3

2

3\times(15-y)=2\times(20-y)3×(15−y)=2×(20−y)

45-3y=40-2y45−3y=40−2y

-3y+2y=40-45−3y+2y=40−45

-y=-5−y=−5

y=5y=5

To find Value of x we should know sum of all frequencies.

Let say sum of all frequencies be f

so, steps to find value of x

Sum of all frequencies = f

6 +7 + y + 15 + 10 + x = f

6 + 7 + 5 + 15 +10 + x = f

x + 43 = f

x = f - 43

Therefore, Value of x is (sum of frequencies) - 43.