SOLUTION: It was found out that average intake of medicine in a group of patients is 110mg with standard deviation 10mg per week.
(a) a patient is chosen at random. Find the probability tha
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Question 535147: It was found out that average intake of medicine in a group of patients is 110mg with standard deviation 10mg per week.
(a) a patient is chosen at random. Find the probability that his/her intake is in betweem 100mg and 115mg per week.
(b) a sample of 3 patients is chosen at random. Find the probability that the sample mean (average) intake exceeds 115mg per week.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
It was found out that average intake of medicine in a group of patients is 110mg with standard deviation 10mg per week.
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(a) a patient is chosen at random. Find the probability that his/her intake is in betweem 100mg and 115mg per week.
z(100) = (100-110)/10 = -1
z(115) = (115-110)/10 = 1/2
P(100 < x < 115) = P(-1< z <1/2) = 0.5328
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(b) a sample of 3 patients is chosen at random. Find the probability that the sample mean (average) intake exceeds 115mg per week.
t(115) = (115-110)/(10/sqrt(3)) = 5/(10/sqrt(3)) = 0.8660
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P(x > 115) = P(t > 0.8660 when df = 2) = 0.2388
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Cheers,
Stan H.
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