Question

find the missing frequency xyz when it is given y:z=4:3 and mean =50

Class Interval. frequency
0-20. 17
20-40. x
40-60. y
60-80. z
80-100. 19​



plz answer correctly

Answer 1

Step-by-step explanation:

The mean of the following frequency table 50. But the frequencies f1 and f2 in class 20 - 40 and 60 - 80 are missing. Find the missing frequencies.Class:0 ...

Class f1 Frequency xi Mid - values ui = xi - A/h fiui 0 - 20 17 10 - 2 - 34 20 - 40 f1 30 - 1 - f1 40 - 60 32 50 0 0 60 - 80 f2 70 1 - f2 80 - 100 19 90 ...

Answer 2

Step-by-step explanation:

Solution:

Class- Interval Mid value(V) Frequency(F=F_{i}F=F

i

) V F

0-20 10 17 170

20-40 30 f 1 30 f 1

40-60 50 f 2 50 f 2

60-80 70 f 3 70 f 3

80-100 90 19 1710

SUM(F)=120 SUM(VF)=Mean=50

1. 17+f 1+f 2+f 3+19= 120

→ f 1+f 2+f 3

= 120 - 36=84

→ f 2 : f 3 = 4 : 3

So,Let, f 2 and f 3 = 4 k and 3 k

→f 1 + 4 k + 3 k= 84

→→f 1 + 7 k= 84------(1)

2. 170 + 30 f 1 + 50 f 2 + 70 f 3 + 1710 = 50×120

→170 + 30 f 1 + 50 × 4 k + 70 × 3 k + 1710= 6000

→170 + 30 f 1 + 200 k + 210 k + 1710= 6000

→30 f 1 + 410 k= 4120

→→ 3 f 1 + 41 k = 412 ----(2)

→→f 1 + 7 k= 84------(1)

=3 × equation(1) - equation (2)

⇒21 k - 41 k = 252- 412

⇒- 20 k = - 160

⇒k=8

Substituting the value of k in equation (2)

→ f 1 +56=84

→ f 1 = 84-56

→ f 1= 28

→f 2 = 4 × 8= 32

→f 3 = 3 × 8 = 24