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Unformatted text preview: 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 36 344. Mean and variance for exercise 318 E( X ) = 3 ∞ ⎛1⎞ 3 ∞ ⎛1⎞ 1 x⎜ ⎟ = ∑ x ⎜ ⎟ = ∑ ⎝ 4 ⎠ 4 x=1 ⎝ 4 ⎠ 3 4 x =0 h( a ) = ∑ a x =
x =1 ∞ x x The result uses a formula for the sum of an infinite series. The formula can be derived from the fact that the series to sum is the derivative of a 1− a with respect to a. For the variance, another formula can be derived from the second derivative of h(a) with respect to a. Calculate from this formula 3 ∞ ⎛1⎞ 3 ∞ ⎛1⎞ 5 E( X ) = ∑ x 2 ⎜ ⎟ = ∑ x 2 ⎜ ⎟ = 4 x =0 ⎝ 4 ⎠ 4 x =1 ⎝ 4 ⎠ 9 514 2 2 Then V ( X ) = E ( X ) − [E ( X ) ] = − = 999
2 x x 345. Mean and variance for random variable in exercise 319 μ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2)
= 0(0.033) + 1(0.364) + 2(0.603) = 1.57
V ( X ) = 02 f (0) + 12 f (1) + 22 f (2) − μ 2 = 0(0.033) + 1(0.364) + 4(0.603) − 1.57 2 = 0.3111
346. Mean and variance for exercise 320 μ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3)
= 0(8 ×10−6 ) + 1(0.0012) + 2(0.0576) + 3(0.9412) = 2.940008 V ( X ) = 02 f (0) + 12 f (1) + 22 f (2) + 32 f (3) − μ 2 = 0.05876096 347. Determine x where range is [0,1,2,3,x] and mean is 6. μ = E ( X ) = 6 = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + xf ( x) 6 = 0(0.2) + 1(0.2) + 2(0.2) + 3(0.2) + x(0.2) 6 = 1.2 + 0.2 x 4.8 = 0.2 x x = 24
37 348. (a) F(0)=0.17 Nickel Charge: X 0 2 3 4 CDF 0.17 0.17+0.35=0.52 0.17+0.35+0.33=0.85 0.17+0.35+0.33+0.15=1 (b)E(X) = 0*0.17+2*0.35+3*0.33+4*0.15=2.29 V(X) = ∑ f ( x )( x − μ )
i =1 i i 4 2 = 1.5259 349. X = number of computers that vote for a left roll when a right roll is appropriate. µ = E(X)=0*f(0)+1*f(1)+2*f(2)+3*f(3)+4*f(4) = 0+0.0003999+2*5.999*108+3*3.9996*1012+4*1*1016= 0.0004 V(X)= ∑ f ( x )( x − μ )
i =1 i i 4 5 2 = 0.00039996 350. µ=E(X)=350*0.06+450*0.1+550*0.47+650*0.37=565 V(X)= σ= ∑ f ( x )( x − μ )
i =1 i 2 =6875 V ( X ) =82.92 351. (a) Transaction New order Payment Order status Delivery Stock level total Frequency 43 44 4 5 4 100
2 µ= Selects: X 23 4.2 11.4 130 0 f(X) 0.43 0.44 0.04 0.05 0.04 E(X) = 23*0.43+4.2*0.44+11.4*0.04+130*0.05+0*0.04 =18.694 V(X) = ∑ f ( x )( x − μ )
i =1 i 5 = 735.964 σ = V ( X ) = 27.1287 (b) Transaction New order Payment Order status Delivery Stock level total Frequency 43 44 4 5 4 100 All operation: X 23+11+12=46 4.2+3+1+0.6=8.8 11.4+0.6=12 130+120+10=260 0+1=1 f(X) 0.43 0.44 0.04 0.05 0.04 µ = E(X) = 46*0.43+8.8*0.44+12*0.04+260*0.05+1*0.04=37.172 38 V(X) = ∑ f ( x )( x − μ )
i =1 i 5 2 =2947.996 σ = V ( X ) = 54.2955 Section 35 352. 353. 354. E(X) = (0+100)/2 = 50, V(X) = [(1000+1)21]/12 = 850 E(X) = (3+1)/2 = 2, V(X) = [(31+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 ⎝ ⎦ 355. ⎛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 6666.67 356. 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 357. 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 = 14.36 358. 359. 39 ...
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This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.
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