the following data gives the number of seeds germination out of 10 damp filters for 80 set of seeds. fit a binomial distribution to these data: X-0,1,2,3,4,5,6,7,8,9,10 F-6,20,28,12,8,6,0,0,0,0,0.
Answers
Answer:
Explanation:
Here the random variable X denotes the number of seeds germinating out of a set of 10 seeds. The total number of trials n = 10.
The mean of the given data
X= 0*6+1*20+2*28+3*12 4*8+5*6/80
=174/80
=2.175
x 0 1 2 3 4 5 6 7 8 9 10 f 6 20 28 12 8 6 0 0 0 0 0
to Sol. Given n = 10, N=E 4=6+20+28+12+8+6=80
Also mean
i=0
10 Ix; ISO fi 0+20+56+36+32+30 = _ - 174 - a175 10 80 80 E 4 it 0
But mean of the binomial distribution = np = 2.175 q= -p 1 - 0.2175 = 0.7825 Hence the binomial distribution to be fitted is = N (q + pY' = 80 (0.7825 + 0.2175)19 = 80.19; (0.7825)19+ 80.10C1(0.7825)9(0.2175)1 + 80."C2(0.7825)9(0.2175)2+ + 80.19C9(0.7825)1(0.2175)9 + 80.1°C10(0.2175)19 = 6.885 + 19.13 + 23.94 + +0.0007+0.00002 The successive terms in the above expansion give the expected or theoretical frequencies, which are
p= 2175 10 - 0.2176
x 0 1 2 3 4 5 6 7 8 9 10 f 6.9 19.1 24 17.8 8.6 2.9 0.7 0.1 0 0 0