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Unformatted text preview: Assignment 3 Due date: 21 October 2009 Total number of points: 32 Q 1. A medical research team wished to evaluate a proposed screening test for Alzheimer’s disease. The test was given to a random sample of 450 patients with Alzheimer’s disease, in 436 cases the test result was positive. Also, the test was given to a random sample of 500 patients without the disease, only in 5 cases the result was positive. It is known that in the US 11.3% of the population aged 65 and over have Alzheimer’s disease. Find the probability that a person has the disease given that the test was positive. Solution to Q1: A test positive, D a person has disease. Given: P ( A  D ) = 436 450 , P ( A  D c ) = 5 500 , P ( D ) = 0 . 113. To find: P ( D  A ) P ( D  A ) = P ( A and D ) P ( A ) We have P ( A and D ) = P ( A  D ) P ( D ) = 436 450 . 113 = 0 . 1094844 Compute first P ( A ) = P ( A  D ) P ( D ) + P ( A  D c ) P ( D c ) = 436 450 . 113 + 5 500 (1 . 113) = 0 . 1183544 The final answer is P ( D  A ) = 0 . 925 Marking scheme for Q1: Total  2 points Q 2. (3.43) A certain drug causes kidney damage in 1% of patients. Suppose the drug is to be tested on 50 patients. Find the probability that (a) none of the patients will experience kidney damage; (b) one or more of the patients will experience kidney damage....
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This document was uploaded on 01/25/2011.
 Spring '08

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