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
0
.
113 = 0
.
1094844
Compute first
P
(
A
) =
P
(
A

D
)
P
(
D
) +
P
(
A

D
c
)
P
(
D
c
) =
436
450
0
.
113 +
5
500
(1
−
0
.
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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 Spring '09
 Kulik
 Normal Distribution, Standard Deviation, Probability theory, Correct Answer, Self number, Kidney damage

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