Find the missing frequencies (f1,f2,f3) in the following frequency distribution when it is given that f2 : f3 = 4 : 3 and mean = 50 CLASS INTERVAL0-20 20-4040-6060-8080-100 FREQUENCY 17f1f2f319Total - 120
Answers
Solution:
Class- Interval Mid value(M) Frequency(F) M 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(M F)=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, 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)
→→→3 × equation(1) - equation (2)
21 k - 41 k = 252- 412
- 20 k = - 160
Dividing both sides by 20, we get
k=8
Substituting the value of k in equation (2)
3 f 1 = 412 - 41 × 8
→ 3 f 1 = 412 - 328
→ 3 f 1= 84
Dividing both sides by , 3
f 1 = 28
f 2 = 4 × 8= 32
f 3 = 3 × 8 = 24