Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
STA 261 – Winter 2011 – Practice Problems Week 1 - Solution 7.11 P(|Y – μ| ≤ 2) = P(–1.5 ≤ Z ≤ 1.5) = 1 – 2P(Z > 1.5) = 1 – 2(.0668) = .8664. 7.12 From Ex. 7.11, we require P(|Y – μ| ≤ 1) = P(–.25 n ≤ Z ≤ .25 n ) = .90. This will be solved by taking .25 n = 1.645, so n = 43.296. Hence, sample 44 trees. 7.15 Using Theorems 6.3 and 7.1: a. E ( X Y ) 1 2 . b. V ( X Y ) 12 / m 22 / n . c. It is required that P(| X Y (1 2 ) | 1) = .95. Using the result in part b for standardization with n = m, 12 2, and 22 2.5 , we obtain n = 17.29. Thus, the two sample sizes should be at least 18. 7.20 a. Using the fact that the chi–square distribution is a special case of the gamma distribution, E(U) = ν, V(U) = 2ν. b. Using Theorem 7.3 and the result from part a: 2 2 E ( S 2 ) n1 E ( n21 S 2 ) n1 ( n 1) σ2. V (S 2 ) V( 2 n 1 2 n 1 2 S2) 2(n 1) 2σ /(n – 1). 2 n 1 2 4 7.29 By Definition 7.3, Y = (W1 / 1 ) (W2 / 2 ) has an F distribution with ν1 numerator and ν2 denominator degrees of freedom. Therefore, U = 1/Y = (W2 / 2 ) (W1 / 1 ) has an F distribution with ν2 numerator and ν1 denominator degrees of freedom. 7.30 a. E(Z) = 0, E(Z2) = V(Z) + [E(Z)]2 = 1. b. This is very similar to Ex. 5.86, part a. Using that result, it is clear that i. E(T) = 0 ii. V(T) = E(T2) = νE(Z2/Y) = ν/(ν–2), ν > 2. 7.33 Define T = Z / W / as in Definition 7.2. Then, T 2 Z 2 /(W / ) . Since Z2 has a chi– square distribution with 1 degree of freedom, and Z and W are independent, T2 has an F distribution with 1 numerator and ν denominator degrees of freedom. 1 7.34 This exercise is very similar to Ex. 5.86, part b. Using that result, is can be shown that a. E ( F ) 12 E (W1 ) E (W21 ) 12 21 2 2 /( 2 2) , ν2 > 2. E (W )E(W ) = ( 2) – = 2 ( 2)/ ( 2) ( b. V ( F ) E ( F 2 ) [ E ( F )]2 2 2 1 2 2 1 2 2 1 v2 2 2 2 2 2 2 1 v2 2 2 2 1 ( 2 2 )( 2 4 ) 1 2 1 2 1 2 2 4) , ν2 > 4. 7.35 Using the result from Ex. 7.34, a. E(F) = 70/(70–2) = 1.029. b. V(F) = [2(70)2(118)]/[50(68)2(66)] = .076 c. Note that the value 3 is (3 – 1.029)/ .076 = 7.15 standard deviations above this mean. This represents and unlikely value. 7.37 a. By Theorem 7.2, χ2 with 5 degrees of freedom. b. By Theorem 7.3, χ2 with 4 degrees of freedom (recall that σ2 = 1). 5 c. Since Y62 is distributed as χ2 with 1 degrees of freedom, and i 1 (Yi Y ) 2 and Y62 are independent, the distribution of W + U is χ2 with 4 + 1 = 5 degrees of freedom. 7.38 a. By Definition 7.2, t–distribution with 5 degrees of freedom. b. By Definition 7.2, t–distribution with 4 degrees of freedom. c. Y follows a normal distribution with μ = 0, σ2 = 1/5. So, 5Y is standard normal and 2 5Y is chi–square with 1 degree of freedom. Therefore, 5Y 2 + Y62 has a chi–square distribution with 2 degrees of freedom (the two random variables are independent). Now, the quotient 2(5Y 2 Y62 ) / U [(5Y 2 Y62 ) / 2] [U / 4] has an F-distribution with 2 numerator and 4 denominator degrees of freedom. Note: we have assumed that Y and U are independent (as in Theorem 7.3). 7.43 Let Y denote the mean height and σ = 2.5 inches. By the Central Limit Theorem, 5(10 ) ) P(| Y | .5) P( .5 Y .5) P( .52(.10 5 Z 2.5 ) P ( 2 Z 2 ) = .9544. 7.53 a. With these given values for μ and σ, note that the value 0 has a z–score of (0 – 12)/9 = 1.33. This is not considered extreme, and yet this is the smallest possible value for CO concentration in air. So, a normal distribution is not possible for these measurements. b. Y is approximately normal: P(Y 14) P( Z 100 (91412 ) ) P( Z 2.22) = .0132. 2 7.55 a. i. We assume that we have a random sample ii. Note that the standard deviation for the sample mean is .8/ 30 = .146. The endpoints of the interval (1, 5) are substantially beyond 3 standard deviations from the mean. Thus, the probability is approximately 1. b. Let Yi denote the downtime for day i, i = 1, 2, …, 30. Then, 30 P( i 1Yi 115) P(Y 3.833) P( Z 30 ( 3..88334 ) ) P( Z 1.14) = .1271. 7.59 Using the result of Ex. 7.58, we have that n = 50, σ1 = σ2 = 2 and μ1 = μ2. Let X denote the mean time for operator A and let Y denote the mean time for operator B (both measured in seconds) Then, operator A will get the job if X – Y < –1. This probability is P( X – Y < –1) ≈ P Z ( 441) / 50 P( Z 2.5) = .0062. 7.73 Let Y = # the show up for a flight. Then, Y is binomial with n = 160 and p = .95. The probability of interest is P(Y ≤ 155), which gives the probability that the airline will be able to accommodate all passengers. Using the normal approximation, this is (.95 ) P(Y 155) P( Z 155160.5(.160 ) P( Z 1.27) .8980 . 95 )(.05 ) 7.75 Let Y = # the favor the bond issue. Then, the probability of interest is P Yn p .06 P .06 Yn p .06 P .2.06(.8 ) Z ..06 P( 1.2 Z 1.2) = .7698. 2 (.8 ) 64 64 7.77 (Similar to Ex. 7.76). Here, we must solve .15 pq / n = z.01 = 2.33. Using p = 1/2, we find that n = 60.32, so 61 customers should be sampled. 7.83 a. Let Y = # that turn right. Then, Y is binomial with n = 50 and p = 1/3. Using the applet, P(Y ≤ 15) = .36897. b. Let Y = # that turn (left or right). Then, Y is binomial with n = 50 and p = 2/3. Using the applet, P(Y ≥ (2/3)50) = P(Y ≥ 33.333) = P(Y ≥ 34) = .48679. 7.87 Let X = waiting time over a 2–day period. Then, X is exponential with β = 10 minutes. Let Y = # of customers whose waiting times is greater than 10 minutes. Then, Y is binomial with n = 100 and p is given by p 101 e y / 10 dy e 1 = .3679. Thus, P(Y 50) P( Z 10 50 100 (.3697 ) 1000 (.3679 )(.6321) P( Z 2.636) = .0041. 3