Math, asked by nasrin67, 5 months ago

The mean of frequency is 53. Find f1 and f2


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Answered by BTSARMYAARTI
0

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

Answered by biligiri
0

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

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