# STAT230 Study Guide - Midterm Guide: Blood Test, Random Variable, Moment-Generating Function

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1. Suppose that npeople take a blood test for a disease, where each person has probability pof having

the disease, independent of other people. To save time and money, blood samples from kpeople are

pooled and analyzed together. If none of the kpersons has the disease then the test will be negative,

but otherwise it will be positive. If the pooled test is positive then each of the kpersons is tested

separately (so k+ 1 tests are done in that case).

[3] (a) Let Xbe the number of tests required for a group of kpeople, then Xtakes possible values at 1

or k+ 1, depending on the outcome of the pooled test. Show that

E(X) = k+ 1 −k(1 −p)k.

Soln: f(1) = P(X= 1) = (1 −p)kand f(k+ 1) = P(X=k+ 1) = 1 −(1 −p)k.

E(X) = 1 ×f(1) + (k+ 1)f(k+ 1) = (1 −p)k+ (k+ 1)[1 −(1 −p)k] = k+ 1 −k(1 −p)k.

[2] (b) What is the expected number of tests required for n/k groups of kpeople each? (Note: the total

number of people involved is n)

Soln: The expected number of tests is (n/k)E(X) = n+n/k −n(1 −p)k

[1] (c) Show that if pis small, the expected number of tests in part (b) is approximately n(kp +k−1),

and is minimized for k=p−1/2.

Soln: Noting that (1 −p)k= 1 −kp +k(k+ 1)p2+···, we have (1 −p)k.

= 1 −kp when pis small.

So

n+n/k −n(1 −p)k.

=n+n/k −n(1 −kp) = n(kp +k−1).

Using the fact that a2+b2≥2ab, and the equality holds when a=b, we have

n(kp +k−1)≥2nqkp ×√k−1= 2n√p ,

and the minimum value 2n√pis achieved when kp =k−1, i.e., k=p−1/2.