Given that XÂ 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.
