Unformatted text preview: 327. X = number of components that meet specifications P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663 P(X=3) = (0.95)(0.98)(0.99) = 0.92169 Section 33 328. x<0 ⎫ ⎧ 0, ⎪1 / 3 0 ≤ x < 1.5⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨2 / 3 1.5 ≤ x < 2⎬ ⎪5 / 6 2 ≤ x < 3 ⎪ ⎪ ⎪ ⎪ 3≤ x ⎪ ⎩1 ⎭ f X (0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3
where f X (1.5) = P ( X = 1.5) = 1 / 3 f X (2) = 1 / 6 f X (3) = 1 / 6 329. x < −2 ⎫ ⎧ 0, ⎪1 / 8 − 2 ≤ x < −1⎪ ⎪ ⎪ ⎪3 / 8 − 1 ≤ x < 0 ⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ 0 ≤ x <1 ⎪ ⎪5 / 8 ⎪7 / 8 1≤ x < 2 ⎪ ⎪ ⎪ ⎪1 2≤ x ⎪ ⎩ ⎭ f X (−2) = 1 / 8 f X (−1) = 2 / 8
where f X ( 0) = 2 / 8 f X (1) = 2 / 8 f X ( 2) = 1 / 8 a) P(X ≤ 1.25) = 7/8 b) P(X ≤ 2.2) = 1 c) P(1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4 d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8 330. ⎧ 0, ⎪ 1 / 25 ⎪ ⎪ ⎪ 4 / 25 F ( x) = ⎨ ⎪ 9 / 25 ⎪16 / 25 ⎪ ⎪1 ⎩ x<0 ⎫ 0 ≤ x < 1⎪ ⎪ ⎪ 1 ≤ x < 2⎪ ⎬ 2 ≤ x < 3⎪ 3 ≤ x < 4⎪ ⎪ 4≤ x ⎪ ⎭ f X (0) = 1 / 25 f X (1) = 3 / 25
where f X (2) = 5 / 25 f X (3) = 7 / 25 f X (4) = 9 / 25 a) P(X < 1.5) = 4/25 b) P(X ≤ 3) = 16/25 c) P(X > 2) = 1 − P(X ≤ 2) = 1 − 9/25 = 16/25 d) P(1 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 1) = 9/25 − 4/25 = 5/25 = 1/5 331. 33 x<0 ⎫ ⎧ 0, ⎪0.008, 0 ≤ x < 1⎪ ⎪ ⎪ ⎪ ⎪ F ( x) = ⎨0.104, 1 ≤ x < 2 ⎬ ⎪0.488, 2 ≤ x < 3⎪ ⎪ ⎪ ⎪ 1, 3≤ x ⎪ ⎩ ⎭
. f (0) = 0.2 3 = 0.008,
where f (1) = 3(0.2)(0.2)(0.8) = 0.096, f (2) = 3(0.2)(0.8)(0.8) = 0.384, f (3) = (0.8) 3 = 0.512, 332. x<0 ⎫ ⎧ 0, ⎪ 0.9996, 0 ≤ x < 1 ⎪ ⎪ ⎪ ⎪ ⎪ F ( x ) = ⎨ 0.9999, 1 ≤ x < 3 ⎬ ⎪0.99999, 3 ≤ x < 4 ⎪ ⎪ ⎪ 4≤ x ⎪ ⎪ 1, ⎩ ⎭ . where f (0) = 0.99994 = 0.9996, f (1) = 4(0.99993 )(0.0001) = 0.0003999, f (2) = 5.999 *10−8 , f (3) = 3.9996 *10−12 , f (4) = 1*10−16 333. x < 10 ⎫ ⎧ 0, ⎪0.2, 10 ≤ x < 25 ⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ ⎪0.5, 25 ≤ x < 50⎪ ⎪ 1, 50 ≤ x ⎪ ⎩ ⎭
where P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 34 334. x <1 ⎫ ⎧ 0, ⎪ 0.1, 1 ≤ x < 5 ⎪ ⎪ ⎪ F ( x) = ⎨ ⎬ ⎪0.7, 5 ≤ x < 10⎪ ⎪ 1, 10 ≤ x ⎪ ⎩ ⎭
where 335. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.5, f(3) = 0.5 a) P(X ≤ 3) = 1 b) P(X ≤ 2) = 0.5 c) P(1 ≤ X ≤ 2) = P(X=1) = 0.5 d) P(X>2) = 1 − P(X≤2) = 0.5 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.7, f(4) = 0.2, f(7) = 0.1 a) P(X ≤ 4) = 0.9 b) P(X > 7) = 0 c) P(X ≤ 5) = 0.9 d) P(X>4) = 0.1 e) P(X≤2) = 0.7 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(10) = 0.25, f(30) = 0.5, f(50) = 0.25 a) P(X≤50) = 1 b) P(X≤40) = 0.75 c) P(40 ≤ X ≤ 60) = P(X=50)=0.25 d) P(X<0) = 0.25 e) P(0≤X<10) = 0 f) P(−10<X<10) = 0 336. 337. 35 338. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1 a) P(X≤1/18) = 0 b) P(X≤1/4) = 0.9 c) P(X≤5/16) = 0.9 d) P(X>1/4) = 0.1 e) P(X≤1/2) = 1 Section 34 339. Mean and Variance μ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + 4 f (4)
= 0(0.2) + 1(0.2) + 2(0.2) + 3(0.2) + 4(0.2) = 2 V ( X ) = 0 2 f (0) + 12 f (1) + 2 2 f (2) + 3 2 f (3) + 4 2 f (4) − μ 2 = 0(0.2) + 1(0.2) + 4(0.2) + 9(0.2) + 16(0.2) − 2 2 = 2
3 40. Mean and Variance for random variable in exercise 314 μ = E ( X ) = 0 f (0) + 1.5 f (1.5) + 2 f (2) + 3 f (3)
= 0(1 / 3) + 1.5(1 / 3) + 2(1 / 6) + 3(1 / 6) = 1.333 V ( X ) = 0 2 f (0) + 1.5 2 f (1) + 2 2 f (2) + 3 2 f (3) − μ 2 = 0(1 / 3) + 2.25(1 / 3) + 4(1 / 6) + 9(1 / 6) − 1.333 2 = 1.139
341. Determine E(X) and V(X) for random variable in exercise 315 . μ = E ( X ) = −2 f ( −2) − 1 f (−1) + 0 f (0) + 1 f (1) + 2 f (2)
= −2(1 / 8) − 1( 2 / 8) + 0(2 / 8) + 1( 2 / 8) + 2(1 / 8) = 0 V ( X ) = −2 2 f (−2) − 12 f (−1) + 0 2 f (0) + 12 f (1) + 2 2 f (2) − μ 2 = 4(1 / 8) + 1(2 / 8) + 0(2 / 8) + 1(2 / 8) + 4(1 / 8) − 0 2 = 1.5
342. Determine E(X) and V(X) for random variable in exercise 316 μ = E ( X ) = 1 f (1) + 2 f (2) + 3 f (3) = 1(0.5714286) + 2(0.2857143) + 3(0.1428571) = 1.571429 V ( X ) = 12 f (1) + 22 f (2) + 32 f (3) + − μ 2 = 1.428571
343. Mean and variance for exercise 317 μ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + 4 f (4)
= 0(0.04) + 1(0.12) + 2(0.2) + 3(0.28) + 4(0.36) = 2.8 V ( X ) = 0 2 f (0) + 12 f (1) + 2 2 f (2) + 3 2 f (3) + 4 2 f (4) − μ 2 = 0(0.04) + 1(0.12) + 4(0.2) + 9(0.28) + 16(0.36) − 2.8 2 = 1.36 36 ...
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 Spring '10
 Feng
 Integers, Standard Deviation, Probability theory, Million, 10000000

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