Answer:
The probability that more than half of them have Type A blood in the sample of 8 randomly chosen donors is P(X>4)=0.1738.
Step-by-step explanation:
This can be modeled as a binomial random variable with n=8 and p=0.4.
The probability that k individuals in the sample have Type A blood can be calculated as:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{8}{k} 0.4^{k} 0.6^{8-k}\\\\\\[/tex]
Then, we can calculate the probability that more than 8/2=4 have Type A blood as:
[tex]P(X>4)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\\\\\P(x=5) = \dbinom{8}{5} p^{5}(1-p)^{3}=56*0.0102*0.216=0.1239\\\\\\P(x=6) = \dbinom{8}{6} p^{6}(1-p)^{2}=28*0.0041*0.36=0.0413\\\\\\P(x=7) = \dbinom{8}{7} p^{7}(1-p)^{1}=8*0.0016*0.6=0.0079\\\\\\P(x=8) = \dbinom{8}{8} p^{8}(1-p)^{0}=1*0.0007*1=0.0007\\\\\\\\P(X>4)=0.1239+0.0413+0.0079+0.0007=0.1738[/tex]
