Given that X be the number of subjects who test positive for the disease out of the 30 healthy subjects used for the test.
The probability of success, i.e. the probability that a healthy subject tests positive is given as 2% = 0.02
Part A:
The probability that all 30 subjects will appropriately test as not being infected, that is the probability that none of the healthy subjects will test positive is given by:
[tex]P(X)={ ^nC_xp^x(1-p)^{n-x}} \\ \\ P(0)={ ^{30}C_0(0.02)^0(1-0.02)^{30-0}} \\ \\ =1(1)(0.98)^{30}=0.5455 [/tex]
Part B:
The mean of a binomial distribution is given by
[tex]\mu=np \\ \\ =30(0.02) \\ \\ =0.6[/tex]
The standard deviation is given by:
[tex]\sigma=\sqrt{np(1-p)} \\ \\ =\sqrt{30(0.02)(1-0.02)} \\ \\ =\sqrt{30(0.02)(0.98)}=\sqrt{0.588} \\ \\ =0.7668[/tex]
Part C:
This test will not be a trusted test in the field of medicine as it has a standard deviation higher than the mean. The testing method will not be consistent in determining the infection of hepatitis.