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Alaboratorybloodtestis99%effectiveindetectingacertaindiseasewhenitisinfactpresent.How...
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a “false positive” result for 1 % of the healthy persons tested. That is, if a healthy person is tested, then with probability 0.01, the test will say he has the disease. If 0.5 % of the population actually has the disease, what is the probability a person has the disease given that his test result is positive?
In a statistics class, the average grade on the final examination was 75 with a standard deviation of 5. The distribution of the data was found to be fairly symmetric.
(i) At least what percentage of the students received grades between 50 and 100? (ii) Determine an interval for the grades that will be true for at least 70% of the students.
(iii) At least what percentage of the students received grades between 50 and 85?
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In a statistics class, the average grade on the final examination was 75 with a standard deviation of 5. The distribution of the data was found to be fairly symmetric.
(i) At least what percentage of the students received grades between 50 and 100? (ii) Determine an interval for the grades that will be true for at least 70% of the students.
(iii) At least what percentage of the students received grades between 50 and 85?
宏观经济概率问题 展开
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(1)
A: has disease (A_c: no disease)
B: test is positive (B_c: test is negative)
P(B|A)=0.99
P(B|A_c)=0.01
P(A)=0.005 => P(A_c)=0.995
P(A|B)= P(B|A)*P(A)/[P(B|A)*P(A)+P(B|A_c)*P(A_c)] (Bayes' theorem)
=0.99*0.005/[0.99*0.005+0.01*0.995] = 33.22%
(2)
If it's a big sample, I think you can pretty much assume it's normally distributed. (given distribution is symmetric). So all you need to do is to standadize the normal using Z=(x-75)/5 and check the normal table to get the answers.
A: has disease (A_c: no disease)
B: test is positive (B_c: test is negative)
P(B|A)=0.99
P(B|A_c)=0.01
P(A)=0.005 => P(A_c)=0.995
P(A|B)= P(B|A)*P(A)/[P(B|A)*P(A)+P(B|A_c)*P(A_c)] (Bayes' theorem)
=0.99*0.005/[0.99*0.005+0.01*0.995] = 33.22%
(2)
If it's a big sample, I think you can pretty much assume it's normally distributed. (given distribution is symmetric). So all you need to do is to standadize the normal using Z=(x-75)/5 and check the normal table to get the answers.
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