SOLUTION: suppose that in a certain area, past experience indicates that the probability of a person with fever will be positive for malaria is 0.7. Consider 3 random selected patients (with

Algebra ->  Probability-and-statistics -> SOLUTION: suppose that in a certain area, past experience indicates that the probability of a person with fever will be positive for malaria is 0.7. Consider 3 random selected patients (with      Log On


   

Question 1186956: suppose that in a certain area, past experience indicates that the probability of a person with fever will be positive for malaria is 0.7. Consider 3 random selected patients (with high fever) in that same area.
a. What is the probability that no patient will be positive for malaria?
b. What is the probability that exactly one patient will be positive for malaria?
c. What is the probability that exactly two of the patients will be positive for malaria?
d. what is the probability that all patients will be positive for malaria?
e. Find the mean and standard deviation of the probability distribution given above.
Answer by ikleyn(52898) About Me  (Show Source):
You can put this solution on YOUR website!
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suppose that in a certain area, past experience indicates that the probability of a person with fever
will be positive for malaria is 0.7. Consider 3 random selected patients (with high fever) in that same area.
a. What is the probability that no patient will be positive for malaria?
b. What is the probability that exactly one patient will be positive for malaria?
c. What is the probability that exactly two of the patients will be positive for malaria?
d. what is the probability that all patients will be positive for malaria?
e. Find the mean and standard deviation of the probability distribution given above.
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(a)  %281-0.7%29%5E3 = 0.3%5E3 = 0.027.                     ANSWER


(b)  C%5B3%5D%5E1%2A0.7%5E1%2A%281-0.7%29%5E2 = 3%2A0.7%2A0.3%5E2 = 0.189.     ANSWER


(c)  C%5B3%5D%5E2%2A0.7%5E2%2A%281-0.7%29%5E1 = 3%2A0.7%5E2%2A0.3 = 0.441.     ANSWER


(d)  0.7%5E3 = 0.343.                                 ANSWER


(e)  mean = n*p = 3*0.7 = 2.1.    standard deviation = sqrt%28n%2Ap%2A%281-p%29%29 = sqrt%283%2A0.7%2A%281-0.7%29%29 = 0.794  (rounded).    ANSWER

Solved.