Practice Problems for Statistical Theory I - Assignment 1 | ST 521, Assignments of Statistics

Download Practice Problems for Statistical Theory I - Assignment 1 | ST 521 and more Assignments Statistics in PDF only on Docsity!

1.1 a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So. for example THT denotes T on Ist, H on 2nd, T on 3rd and T on 4th. There are 24 = 16 such sample points. : b. ‘The mmber of damaged leaves is a nonegative integer. So we might use $ = {0,1,2,...} Vv ¢ We might observe fractions of an hour. So we might use $ = {t:t > 0}, that is, the half infinite interval [0, 00) : : 4. Suppose we weigh the rats in ounces. The weight must be greater than zero so we might use S = (0,00). If we know no 10-day-old rat weighs more than 100 o7., we could use § — (0, 100} \/ © lf nis the number of items in the shipment, then § = {0/n,1/n,.... im 13a. cre AUB & cEeAorzxeB & rE BUA te ANB & ce€AandreB @ cE BNA b. ce AU(BUC) & xe Aorre BUCS re AUBorzeCs re (AUB)UC. (It can similarly be shown that AU (BUC) = (AUC) UB.) teEAN(BNC) > reAandre BandreCs re (ANB)NC. ce. 2E(AUB)*S r¢Aorez¢gBoe creA andre Bs rE ANB ce(AN Be c¢ZANBe z¢gAandr¢~ Bo re A orze BS re AUB. 1.4 a. “Aor Bor both” is AUB. From Theorem 1.2.9b we have P(AUB) = P(A)+P(B)—P(AMB). b. “A or B but not both” is (AN B°)U(BN A°). Thus we have P((ANB)U(BNA)) = P(ANB*)+ P(BN A’) (disjoint union) = [P(A)- P(ANB)]+[P(B)- P(ANB)] (Theorem1.2.9a) P(A) + P(B) - 2P(AN B). At least one of A or BY is AU B. So we get the same answer as in a) wi ‘At most one,of A or B” is (AN B)®, and P((AN B)*) =1~- P(ANB). 1.5 a. ANBNC = {a US. birth results in identical twins that are female} b. PANBNC)=%x4x$d 16 po=(1-u)(l—w), pr=u-w)+w(l-u), po =u, Po=p2 => utw=l p=m > w=1/3. These two equations imply u(1 — u) = 1/3, which has no solution in the real numbers. Thus, the probability assignment is not legitimate. LT a ee b. Plocoring¢points|board is hit) = PGsorng Tpotes beer su) P(board is hit) = ae P(scoring i points board is hit) = ia (S 2 “= GE 16. Therefore, (6 -# - (6-1)? P(scoring i points|board is hit) = which is exactly the probability distribution of Example 1.2.7. 1.13 If A and B are disjoint, P(AU B) = P(A) + P(B) = 1 generally, if A and B are disjoint, then A c B¢ and P(A so A and B cannot be disjoint. + 3 = 15, which is impossible. More ys (B°). But here P(A) > P(B®), the z 2