mont4e_sm_ch03_sec05

mont4e_sm_ch03_sec05 - V(X) = ∑ f ( x )( x − μ ) i =1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V(X) = ∑ f ( x )( x − μ ) i =1 i 5 2 =2947.996 σ = V ( X ) = 54.2955 Section 3-5 3-52. 3-53. 3-54. E(X) = (0+100)/2 = 50, V(X) = [(100-0+1)2-1]/12 = 850 E(X) = (3+1)/2 = 2, V(X) = [(3-1+1)2 -1]/12 = 0.667 X=(1/100)Y, Y = 15, 16, 17, 18, 19. E(X) = (1/100) E(Y) = 2 1 ⎛ 15 + 19 ⎞ ⎜ ⎟ = 0.17 mm 100 ⎝ 2 ⎠ 2 ⎛ 1 ⎞ ⎡ (19 − 15 + 1) − 1⎤ 2 V (X ) = ⎜ ⎟⎢ ⎥ = 0.0002 mm 100 ⎠ ⎣ 12 ⎝ ⎦ 3-55. ⎛1⎞ ⎛1⎞ ⎛1⎞ E ( X ) = 2⎜ ⎟ + 3⎜ ⎟ + 4⎜ ⎟ = 3 ⎝3⎠ ⎝3⎠ ⎝3⎠ in 100 codes the expected number of letters is 300 2 2⎛1⎞ 2⎛1⎞ 2⎛1⎞ 2 V ( X ) = (2 ) ⎜ ⎟ + (3) ⎜ ⎟ + (4 ) ⎜ ⎟ − (3) = 3 ⎝3⎠ ⎝3⎠ ⎝ 3⎠ in 100 codes the variance is 6666.67 3-56. X = 590 + 0.1Y, Y = 0, 1, 2, ..., 9 E(X) = ⎛0+9⎞ 590 + 0.1⎜ ⎟ = 590.45 mm ⎝2⎠ 2 2 ⎡ (9 − 0 + 1) − 1 ⎤ V ( X ) = (0.1) ⎢ ⎥ = 0.0825 12 ⎦ ⎣ mm2 3-57. a = 675, b = 700 (a) µ = E(X) = (a+b)/2= 687.5 V(X) = [(b – a +1)2 – 1]/12= 56.25 (b) a = 75, b = 100 µ = E(X) = (a+b)/2 = 87.5 V(X) = [(b – a + 1)2 – 1]/12= 56.25 The range of values is the same, so the mean shifts by the difference in the two minimums (or maximums) whereas the variance does not change. X is a discrete random variable. X is discrete because it is the number of fields out of 28 that has an error. However, X is not uniform because P(X=0) ≠ P(X=1). The range of Y is 0, 5, 10, ..., 45, E(X) = (0+9)/2 = 4.5 E(Y) = 0(1/10)+5(1/10)+...+45(1/10) = 5[0(0.1) +1(0.1)+ ... +9(0.1)] = 5E(X) = 5(4.5) = 22.5 V(X) = 8.25, V(Y) = 52(8.25) = 206.25, σY = 14.36 3-58. 3-59. 3-9 3-60. E (cX ) = ∑ cxf ( x) = c∑ xf ( x) = cE ( X ) , V (cX ) = ∑ (cx − cμ ) f ( x) = c 2 ∑ ( x − μ ) 2 f ( x) = cV ( X ) 2 x x x x Section 3-6 3-61. A binomial distribution is based on independent trials with two outcomes and a constant probability of success on each trial. a) reasonable b) independence assumption not reasonable c) The probability that the second component fails depends on the failure time of the first component. The binomial distribution is not reasonable. d) not independent trials with constant probability e) probability of a correct answer not constant. f) reasonable g) probability of finding a defect not constant. h) if the fills are independent with a constant probability of an underfill, then the binomial distribution for the number packages underfilled is reasonable. i) because of the bursts, each trial (that consists of sending a bit) is not independent j) not independent trials with constant probability 3-62. (a) P(X≤3) = 0.411 (b) P(X>10) = 1 – 0.9994 = 0.0006 (c) P(X=6) = 0.1091 (d) P(6 ≤X ≤11) = 0.9999 – 0.8042 = 0.1957 (a) P(X≤2) = 0.9298 (b) P(X>8) = 0 (c) P(X=4) = 0.0112 (d) P(5≤X≤7) = 1 - 0.9984 = 0.0016 3-63. 3-64. ⎛10 ⎞ = 5) = ⎜ ⎟0.5 5 (0.5) 5 = 0.2461 ⎜5⎟ ⎝⎠ ⎛10 ⎞ 0 10 ⎛10 ⎞ 1 9 ⎛10 ⎞ 2 8 b) P( X ≤ 2) = ⎜ ⎟0.5 0.5 + ⎜ ⎟0.5 0.5 + ⎜ ⎟0.5 0.5 ⎜2⎟ ⎜1⎟ ⎜0⎟ ⎝⎠ ⎝⎠ ⎝⎠ 10 10 10 = 0.5 + 10(0.5) + 45(0.5) = 0.0547 ⎛10 ⎞ 10 ⎛10 ⎞ 9 1 0 c) P ( X ≥ 9) = ⎜ ⎟0.5 (0.5) + ⎜ ⎟0.5 (0.5) = 0.0107 ⎜10 ⎟ ⎜9⎟ ⎝⎠ ⎝⎠ a) P ( X d) ⎛10 ⎞ ⎛10 ⎞ P(3 ≤ X < 5) = ⎜ ⎟0.530.57 + ⎜ ⎟0.540.56 ⎜4⎟ ⎜3⎟ ⎝⎠ ⎝⎠ 10 = 120(0.5) + 210(0.5)10 = 0.3223 ⎛10 ⎞ 5 = 5) = ⎜ ⎟0.015 (0.99) = 2.40 × 10−8 ⎜5⎟ ⎝⎠ 3-65. a) P ( X 3-10 ...
View Full Document

Ask a homework question - tutors are online