It is estimated that 0.16% of the adult population in this country without tuber
ID: 3440719 • Letter: I
Question
It is estimated that 0.16% of the adult population in this country without tuberculosis is infected with HIV. Compute the probability that a randomly selected adult is infected with HIV. Given that a randomly selected adult is infected with HIV, compute the probability that this adult is co-infected with tuberculosis. Tuberculosis is a rare disease in developing countries. However, it is the most common cause of death in HIV-positive adults living in these countries Consider a country in which 0.045% of its adult population has tuberculosis. It is estimated that 8% of all tuberculosis cases in this country are co-infected with HIV. Compute the probability that a randomly selected adult from this country has both tuberculosis and HIV. P(HIV and Tub) = 0.000036Explanation / Answer
PROBLEM 8.15
are mutually exclusive?
No because the events are choosing for a specific country so the person who is choose can have HIV or tuberculosis or both
are the events independents?
Yes, because HIV is independent of tuberculosis that means, if the person have HIV not dependens of if he has tuberculosis
PROBLEM 8.17
a)
Expected value for geese that dont have a schistosomiasis infection
e(x)= (1-0.05)*5 = 4.75
b)
P(at least one of the geese does not have a schistosomiasis infection) = P(x=1) +P(x=2) +P(x=3) + P(x=4) + P(x=5)
binomial distribution
n=5
p=0.95
q= 0.05
P(x=1) = (5C1) * 0.95^1 * 0.05 ^ (5-1) = 0.000030
P(x=2) = (5C2) * 0.95^2 * 0.05 ^ (5-2) = 0.001128
P(x=3) = (5C3) * 0.95^3 * 0.05 ^ (5-3) = 0.0214
P(x=4) = (5C4) * 0.95^4 * 0.05 ^ (5-4) = 0.2036
P(x=5) = (5C5) * 0.95^5 * 0.05 ^ (5-5) = 0.7738
P( at least one of the feese does not have infection) = 0.7738+0.2036+0.0214+ 0.001128+0.000030
P( at least one of the feese does not have infection) = 0.999958
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