Answer:
Step-by-step explanation:
Let X be the number of persons out of 12 who have RH negative blood.
X is binomial since each person is independent of the other to have Rh negative blood.
n =12
a) What is the probability that exactly 5 of those patients will have Rh negative blood?
P(X=5) =[tex]12C5 (0.26)^5 (1-0.26)^7\\= 0.1143[/tex]
b) What is the probability that at least 3 of them will have Rh negative blood?
[tex]P(X\geq 3) = \Sigma_3^{12} 12Cr (0.26)^r (1-0.26)^{12-r}[/tex]=0.6397
c) What are the expected number and standard deviation of the number of these patients with Rh negative blood?
E(X) = np = [tex]12(0.26) = 3.12[/tex]
Var(x) = npq = [tex]3.12 *0.74[/tex]
Std dev = [tex]\sqrt{Varx} \\=1.5195[/tex]
