Question

In a study of helicopter usage and patient survival, among the 42.237 patients transported by helicopter, 176 of them left the treatment center against medical advice, and the other 42,061 did not leave against medical advice. If 50 of the subjects transported by helicopter are randomly selected without replacement, what is the probability that none of them left the treatment center against medical advice? The probability is (Round to three decimal places as needed.)

Answer 1

a) solution :- from the given data , - Here, =) sample proportion = 176 42237 = 0.004167 0.0042 n= so. - Binomial distribution. P(x) = (2) pt (1-pjh-x so-o P(x=0) = (0) (0.couz (1-0.com)*** = 0.8102 left the treatment The probability center against that none of them the medical advice is (0.8102)

b) solution : From the given data, - Here, The true The true probability of baby being girl q = 0.472 of probability of baby being boy is P: 1-a P=1-9 = 1-0.472 = 0.528 n = 4. =) Foor binomial distribution, P(x-x) = neyx po xq nor. -> P(121) = 1 - PlX :0) I-UCO * 0.528° x 0.472-0 = -0.0496 - 0.990u
Thank you.