If the mean of the distribution is 33.2 and the sum of all frequencies is 100. Find the missing frequencies f1 and f2. Class intervals are 6-14,14-22,22-30,30-38,38-46,46-54,54-62,62-70and the frequencies are 11,21,f1,15,14,8,f2,6
Answers
Step-by-step explanation:
mean of the distribution of 33.2 and the sum of all vacancies hundred then the missing frequency of F1 and F2 are 6262 70 and F2 and F6
Lets calculate values of missing frequencies of f1 and f2
Step-by-step explanation:
Given, Mean = 33.2
Sum of all frequencies = 100
Class Interval Frequency(f) Mid value(m) fm(f*m)
6-14 11 10 110
14-22 21 18 378
22-30 f1 26 26 f1
30-38 15 34 510
38-46 14 42 588
46-54 8 50 400
54-62 f2 58 58 f2
62-70 6 66 396
75 + f1 +f2 2382+26f1 +58f2
Mean = ∑ fm ÷ ∑f
33.2 = ( 2382+ 26 f1 + 58 f2) ÷ (75 + f1 + f2)
2490 + 33.2 f1 + 33.2 f2 = 2382+26 f1 + 58 f2
2490 - 2382 = 26 f1 + 58 f2 -33.2 f1 - 33.2 f2
108 = -7.2 f1 + 24.8 f2 ( Equation 1)
Now Mean =( 2382 +26 f1 +58 f2) ÷ 100
33.2 × 100 = 2382 + 26f1 + 58 f2
3320 = 2382 + 26f1 + 58 f2
938 = 26f1 + 58f2 (Equation 2)
From equation 1 & 2
We get, 1062.4 f2 = 9561.6
f2 = 9
Therefore, By putting value of f2
∑ X = 75 + f1 + f2
100 = 75 + f1 + 9
i.e., f1 = 16
Therefore, Answer is f1 = 16 , f2 = 9