find the missing frequencies f1 f2 and f3 in the following frequency distribution when it is given that f2:f3 and mean=50
Answers
Answer:
f1 = 28, f2 = 32, f3 = 24
Step-by-step explanation:
Since the ratio f2 : f3 = 4 : 3, we can write f2 = 4t and f3 = 3t for some value t.
The total number of observations is 120, so
17 + f1 + f2 + f3 + 19 = 120
=> f1 + 4t + 3t = 120 - 17 - 19
=> f1 + 7t = 84. (*)
The mean is 50, so
( 10 x 17 + 30 f1 + 50 f2 + 70 f3 + 90 x 19 ) / 120 = 50
=> 170 + 30 f1 + 200 t + 210 t + 1710 = 50 x 120 = 6000
=> 17 + 3 f1 + 20 t + 21 t + 171 = 600
=> 3 f1 + 41 t = 600 - 171 - 17 = 412. (**)
Multiplying equation (*) by 3, we have
3 f1 + 21 t = 3 x 84 = 252.
Subtracting this from equation (**), we get
20 t = 412 - 252 = 160 => t = 8.
So f2 = 4t = 32, f3 = 3t = 24, and f1 = 84 - 7t = 84 - 56 = 28.
Answer:
Given frequency =120
⇒17+f
1
+32+f
2
+19=120
⇒f
1
+f
2
=52
Mean =
∑f
i
∑x
i
f
i
⇒mean =
120
(10×17)+(30×f1)+(50×32)+(72×f2)+(90×19)
⇒mean =
120
170+30f
1
+1600+70f
2
+1710
=50
⇒3f
1
+7f
2
=252
⇒3f
1
+7(52−f
1
)=252
⇒f
1
=28
⇒f
2
=24