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

Class Interval. frequency
0-20. 17
20-40

2 thoughts on “find the missing frequency xyz when it is given y:z=4:3 and mean =50<br /><br />Class Interval. frequency<br />0-20. 17<br />20-40”

1. 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

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2. 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 …

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