Answer:
0.8187,0.00115,0.20
Step-by-step explanation:
Given that one out of 5000 individuals in a population carries a certain defective gene.
Hence in the sample of 1000, we find X no of those carrying defective gene will follow a Poisson distribution with mean[tex]= 1/5 = 0.2[/tex]
a)
the probability that none of the sample individuals carries the gene
[tex]=P(x=0)\\= 0.8187[/tex]
b) the probability that more than two of the sample individuals carry the gene
=[tex]P(X>2) = 0.00115[/tex]
c) the mean of the number of sample individuals that carry the gene
0.20