The mean of frequency is 53. Find f1 and f2
plzzz answer
Answers
Step-by-step explanation:
No. of people
(f) xi=2lower limit + upper limitxifi 0-20 15 10 150 20-40 f1 30 30f1 40-60 21 50 1050 60-80 f2 70 70f2 80-100 17 90 1530Σfi=53+f1+f2
Σfi.xi=2730+30f1+70f2
Given:- Σfi=100
⇒53+f1+f2=100
⇒f1+f2=100−53
⇒f1+f2=47⟶eq.(i)
Given:- Mean = ΣfiΣxi.fi=53
Step-by-step explanation:
class xi fi fixi
0 - 20 10 15 150
20 - 40 30 f1 30f1
40 - 60 50 21 1050
60 - 80 70 f2 70f2
80 - 100. 90 17 1530
_______________________________
∑fi = 53 + f1 + f2 ∑fixi = 2730+30f1+70f2
________________________________
given ∑fi = 100
therefore 53 + f1 + f2 = 100
=> f1 + f2 = 47..........................1
mean x = 53
mean = ∑fixi/∑fi
therefore 53 = (2730+30f1+70f2)/100
on cross multiplication we get,
5300 = 2730+30f1+70f2
5300 - 2730 = 30f1+70f2
2570 = 30f1+70f2
257 = 3f1+7f2 [ divide both sides by 10 ]
=> 3f1 + 7f2 = 257..............2
solve equation 1 & 2
3f1+ 7f2 = 257..............2
f1 + f2 = 47 => f2 = 47 - f1......3
substitute equation 3 in 2
3f1 + 7(47 - f1) = 257
3f1 + 329 - 7f1 = 257
- 4f1 = - 72
f1 = 18
f2 = 47 - 18 = 29
Answer