STAT 231 Homework 5 Solutions

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1 Will Landau September 26, 2011 STAT 231 Homework 5 Solutions Exercise 5.1 (Devore 4.11). f. E(X) = xf(x)dx = xf (x)dx = 2 0 x x 2 dx = (1/2) 2 0 x2 dx = [ ] 2 = x g. E(X 2 ) = x2 f(x)dx = x2 F (x)dx = 2 0 x2 x 2 dx = (1/2) 2 0 x3 dx = [ ] 2 x 4 8 = 2. Hence V (X) = E(X 2 ) E 2 (X) = = Also, 0 X = V (X) =.471 h. E(h(X)) = E(X 2 ) = 2 from part g. Exercise 5.2 (Devore 4.13). First, we should get part a: a. cdf: for x > 1, F X (x) = x f(t)dt = x 1 kt 4 dt = [ k 3 t 3] x 1 = k 3x 3 + k 3. We need lim x F (x) = k 3 = 1, so k=3 and F (X) = 1 x 3 if x > 1 and 0 if x 1. d. E(X) = x k 1 x dx = k 4 1 x dx = [ k 3 2 x 2] = 0 + k/2 = k/2= E(X 2 ) = x 2 k 1 x dx = k 4 1 x dx = [ kx 1] = 0 + k = k=3 2 1 V (X) = E(X 2 ) E 2 (X) = k k2 4, X = V = k k2 4 = e. P (X (E(X) X, E(X) + X )) = F (E(X) + X ) F (E(X) X ) = F ( ) F ( ) = F (2.366) F (.634) = ( ) 0 (part a) =.9245 Exercise 5.3 (Devore 4.28). a. Φ(2.17) Φ(0) =.4850 b. Φ(1) Φ(0) =.3413 c. Φ(0) Φ( 2.5) =.4938 d. Φ(2.5) Φ( 2.5) =

2 e. Φ(1.37) =.9147 f. 1 Φ 1.75 =.9599 g. Φ(2) Φ( 1.5) =.9104 h. Φ(2.5) Φ(1.37) =.0791 i. 1 Φ(1.5) =.0668 j. P ( Z 2.5) = P ( 2.5 Z 2.5) = Φ(2.5) Φ( 2.5) =.9876 Exercise 5.4 (Devore 4.29). a. You can find.9838 is found in the 2.1 row and the.04 column of the standard normal table (Table A-3 page A-7) so c = b. Φ(c) Φ(0) =.291 Φ(c) =.791 c =.81 c. P (c Z) = P (clz) = P (Z < c) = Φ(c) = =.8790 c=1.17 d. Φ(c) Φ( c) = Φ(c) (1 Φ(c)) = 2Φ(c) 1 Φ(c) =.9920 c=.97. e..016 = P (c Z ) = 1 P (c Z ) = P ( c Z c) = Φ(c) Φ( c) = 2Φ(c) 1 Φ(c) =.9920 c=2.41 Exercise 5.5 (Devore 4.30). a. Φ(c) =.9100 c 1.34 (Use the table is the entry in the 1.3 row, 0.04 column) b. Since the normal distribution is symmetric with mean 0, 9th percentile = -91st percentile = c. Φ(c) =.7500 c.675 since.7486 and.7517 are in the.67 and.68 entries of the table, respectively. d. 25th percentile = -75th percentile = e. Φ(c) =.06 c (.0594 and.0606 appear as the and entries of the table, respectively). Exercise 5.6 (Devore 4.32). a. P (X 15) = Φ( ) = Φ(0) =.5 b. P (X 17.5) = Φ( 17.5 frm[o] 5.0 ) = Φ(2) =

3 c. P (X 10) = 1 P (X < 10) = 1 Φ( ) = 1 Φ( 4) = = d. P (14 X 18) = P (X 18) P (X < 14) = Φ( ) Φ( ) = Φ() Φ(2.4) Φ(.8) = =.7799 e. P ( X 15 3) = P ( 3 X 15 3) = P (12 X 18) = P (X 18) P (X < 12) = Φ( ) Φ( ) = Φ(2.4) Φ( 2.4) = =.9836 Exercise 5.7 (Devore 4.34). a. P (X >.25) = P (Z >.83) = 1 Φ(.83) = =.7967 b. P (X.10) = Φ( 3.33) =.0004 c. The larges 5% of concentration values fall at or above the 95th percentile. Hence, we need to find the 95th percentile of the normal distribution with mean.30 and standard deviation.06. From the table, the 95th percentile of the standard normal distribution is Thus, if Z N(0, 1), then.95 = P (Z ) = P (.06 Z ) = P (X ). Hence, the 95th percentile is.3987, and the largest 5% of concentration values are the ones at or above Exercise 5.8 (Devore 4.42). Let X be the temperature reading and let Z = X µ N(0, 1). We want P (µ.1 X µ +.1) =.95, which means:.95 = P (.1 X µ.1) = P (.1 X µ.1 ) = P (Z.1.1 ) P (Z < ) = Φ(.1 ) Φ(.1 ) = Φ(.1 ) [1 Φ.1 )] (since the normal density is symmetric about 0) = 2Φ(.1 ) 1 With.95 = 2Φ(.1 ) 1, we get Φ(.1 ) =.975. From the table,.1 means = = 1.96, which Exercise 5.9 (Devore 4.43). If 10% of all resistors exceed ohms, then ohms is the 90th percentile on the normal distribution of resistances: N(µ, 2 ). Hence 3

4 Φ( µ ) =.9. From the normal tables, this means µ = z.9 = 1.28, so µ = 1.28 µ = If 5 % or resistors have a resistance smaller than ohms, then ohms is the 5th percentile on the normal distribution of resistances: N(µ, 2 ). Hence Φ( µ ) =.05. From the normal tables, this means µ = z.05 = 1.64, so µ = 1.64 µ 1.64 = Solving the system of equations: µ = µ 1.64 = we get µ = 10 and =.2 Exercise 5.10 (Devore 4.46). a. P (67 X 75) = P ( 1.00 Z 1.67) =.7938 b. P (70c X 70 + c) = P ( c/3 Z c/3) = Φ(c/3) Φ( c/3) = 2Φ(c/3) 1. Now, if.95 = P (70c X 70 + c), then.95 = 2Φ(c/3) 1 Φ(c/3) =.975 c/3 = 1.96 c=5.88 c. E(# of 10 that are acceptable) = 10?P(a single one is acceptable) = d. p = P (X < 73.84) = P (Z < 1.28) =.9, sop (Y 8) = B(8; 10,.9) =.264 Exercise 5.11 (Devore 4.47). We want.99 = P (W < c 1) =.99, where W N(12, ) is the weight of a random parcel. Standardizing, we get.99 = P (Z < c = Φ( c ). From the table, we get c = z.99 = 2.33 c = c= Exercise 5.12 (Devore 4.53). Remember that for X Binomial(n, p), µ X = np and X = np(1 p). Hence, for p=.5, µ X = 25.5 = 12.5 and X = 25.5 (1.5) = For p=.6, µ X = 25.6 = 15 and X = 25.6 (1.6) = 6. For p=.8, µ X = 25.8 = 20 and X = 25.8 (1.8) = 4. The following tables show the results. a. 4

5 b. c. The comparisons are pretty good, but they decrease as p goes from.5 to.8. In general, the normal approximation improves with large n (for a smoother binomial pdf) and p close to.5 (for a binomial pdf without much skew). Exercise 5.13 (Devore 4.54). p =.10 and n = 200, so µ X = np = 20, and X = np(1 p) = 18 = Let Y N(20, 18). Then, Y approximates X. Using the normal approximation: a. P (X 30) P (Y 30.5) = Φ( ) = Φ(2.47) =.9932 b. P (X < 30) P (Y 29.5) = Φ( ) = Φ(2.24) =.9875 c. P (15 X 25) = P (X 25) P (X 14) P (Y 25.5) P (Y 14.5) = Φ( ) Φ( ) = Φ(1.30) Φ( 1.30) = =.8064 Exercise 5.14 (Devore 4.59). a. E(X) = 1 λ = 1 b. X = 1 λ = 1 c. P (X 4) = F X (4) = 1 e 4 = d. P (2 X 5) = P (X 5) P (X < 2) = F X (5) F X (2) = 1 e 5 (1 e 2 ) = e 2 e 5 = Exercise 5.15 (Devore 4.61). Let X be the rainfall duration. X Exponential(λ = 1/µ = 1/2.725 =.367) a. P(duration is at least 2 hours) = P (X 2) = 1 P (X < 2) = 1 F X (2) = 1 (1 e ) =.480 P(at most 3 hours) = P (X 3) = F X (3) = 1 e =.667 5

6 P(between 2 and 3 hours) = P (2 X 3) = P (X 2) P (X > 3) = P (X 2) (1 P (X 3)) =.480 ( ) (from the previous 2 calculations) =.147 b. µ X = Hence, µ + 2 = 3 = P (X > µ + 2) = 1 P (X µ + 2) = 1 F X (8.175) = 1 (1 e ) =.046 P (X < µ ) = P (X < 0) = F X (0) = 1 e 0 = 0 Exercise 5.16 (Devore 4.69). a. For X t, all components need to last at least t hours. Hence, {X t} = A 1 A 2 A 3 A 4 A 5 b. P (X t) = P (A 1 )P (A 2 )P (A 3 )P (A 4 )P (A 5 ) = P (some component lasts at least t hours) 5 = [1 P (some component lasts less than t hours)] 5 = [1 (1 e.01 t )] 5 = e.05t. Hence, P (X t) = 1 e.05t, so that X Exp(λ =.05) c. By the same reasoning, P (X t) = 1 e nλt, so X has an exponential distribution with parameter nλ. Exercise 5.17 (Devore 4.73). Let X be the lifetime of a random specimen. Then, the cdf of X is F (x) = F (x; α = 2.5, β = 200) = 1 e (x/β)α = 1 e (x/200)2.5 for x 0 and 0 for x < 0. a. At most 250: P (X 250) = F (250) = 1 e (250/200)2.5 = 1 e =.8257 Less than 250: since X is continuous, P (X < 250) = P (X 250) =.8257 More than 300: P (X > 300) = 1 P (X 300) = 1 F (300) = 1 (1 e (300/200)2.5 ) =.0636 b. Between 100 and 250: P (100 X 250) = P (X 250) P (X < 100) =.8257 F (100) (from part a) =.8257 (1 e (100/200)2.5 ) =.8257 (.1620) =

7 c. This question asks for the median m of X. By definition, F (m) =.5, so: 1 e (m/β)α =.5.5 = e (m/β)α ( m log(.5) = β ( ) α m log(2) = β Exercise 5.18 (Devore 5.1). ) α m = β[log(2)] 1/α = 200[log(2)] 1/2.5 = a. P (X = 1, Y = 1) = p(1, 1) =.20 b. P (X 1 and Y 1) = p(0, 0) + p(0, 1) + p(1, 0) + p(1, 1) =.42 c. At least one hose is in use at both islands. P (X 0 and Y 0) = p(1, 1) + p(1, 2) + p(2, 1) + p(2, 2) =.70 d. Sum the row probabilities to get p X (x) =.16,.34,.50 for x = 0, 1, 2. Sum the column probabilities to get p Y (y) =.24,.38,.38 for y = 0, 1, 2. P (X 1) = p X (0) + p X (1) =.50 Exercise 5.19 (Devore 5.3). a. p(1, 1) =.15, the entry in the 1st row and 1st column of the joint probability table. b. P (X 1 = X 2 ) = p(0, 0) + p(1, 1) + p(2, 2) + p(3, 3) = =.40 c. A = {(x 1, x 2 ) x1 2 + x 2 } {(x 1, x 2 ) x x 1 } P (A) = p(2, 0) + p(3, 0) + p(4, 0) + p(3, 1) + p(4, 1) + p(4, 2) + p(0, 2) + p(0, 3) + p(1, 3) =.22 d. exactly 4: P (exactly 4) = p(1, 3) + p(2, 2) + p(3, 1) + p(4, 0) =.17 P (at least 4) = P (exactly 4) + p(4, 1) + p(4, 2) + p(4, 3) + p(3, 2) + p(3, 3) + p(2, 3) =.46 Exercise 5.20 (Devore 5.4). a. P 1 (0) = P (X 1 = 0) = p(0, 0) + p(0, 1) + p(0, 2) + p(0, 3) =.19 P 1 (1) = P (X 1 = 1) = p(1, 0) + p(1, 1) + p(1, 2) + p(1, 3) =.30, etc. 7

8 E(X 1 ) = = 1.7 b. P 2 (0) = P (X 2 = 0) = p(0, 0) + p(1, 0) + p(2, 0) + p(3, 0) + p(4, 0) =.19, etc. c. p(4, 0) = 0 yet p 1 (4) =.12 > 0 and p 2 (0) =.19 > 0, so p(x 1, x 2 ) p 1 (x 1 ) p 2 (x 2 ) for some (x 1, x 2 ). Hence, the two variables are not independent. 8