0020: Engineering Probability and Statistics for Engineers 1 Spring, 2007; Solutions: Homework #5 Chapter 4.4, 4.5 1. Based on data for Question 81, p139. a Let X be the time between aircraft arrivals. X ~ exp(=8) x = E(X) = 1/8 hr. = 7.5 minutes b P(X<5) = 1 e-(8)(5/60) = 0.4866 c P(X>30) = e-(8)(0.5) = 0.0183 d P(X<5 or X>10) = P(X<5) + P(X>10) = 0.4866+ e-(8)(10/60)=0.7502 e P(X>30 | last arrival occurred 30 min. ago) = P(X>30) = 0.0183 This follows from the "Memoryless" property of the Exponential Distribution. Compute the sample and Normal percentiles as shown below: i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (i-0.5)/n 0.025 0.075 0.125 0.175 0.225 0.275 0.325 0.375 0.425 0.475 0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875 0.925 0.975 Xi 47.1 68.1 68.1 90.8 103.6 106 115 126 146.6 229 240 240 278 278 289 289 367 385.9 392 505 Zi -1.96 -1.44 -1.15 -0.93 -0.76 -0.60 -0.45 -0.32 -0.19 -0.06 0.06 0.19 0.32 0.45 0.60 0.76 0.93 1.15 1.44 1.96
86.
A plot of Xi vs. Zi is shown on the next page:
Normal Probability Plot
600 observed data 500 400 300 200 100 -3.00 -2.00 -1.00 0 0.00 z percentile 1.00 2.00 3.00
The plot looks reasonably straight and an assumption of Normality seems to be a plausible one. The intercept (approx. 200+ ??) would be an estimate of the mean and the slope would estimate the std. dev.
Chapter 5 2. a. Since X and Y are independent p(x,y)=pX(x)pY(y). This yields the following: y p(x,y ) 0 1 2 3 4 0 .30 .05 .025 .10 .025 .5 x 1 .18 .03 .015 .015 .06 .3 2 .12 .02 .01 .01 .04 .2 .6 .1 .05 .05 .20 b c d P(X1) and Y1) = p(0,0) + p(1,0) + p(0,1) + p(1,1) = .56 (...OR P(X1) and Y1) = P(X1) * P(Y1) = (.8)(.7) = 0.56) P(X+Y=0) = P(X=0 and Y=0) = p(0,0) = .30 P(X+Y1) = p(0,0) + p(1,0) + p(0,1) = .53 Given fX(x)=e-x , fY(y)=e-y with =1 in both cases, i.e. fX(x)=e-x ; fY(y)=e-y Since X and Y are independent, joint pdf f(x,y)=fX(x) fY(y)=e-x-y. P(X1 and Y1)=P(X1)P(Y1)=(1-e-1)(1-e-1)= 0.400
13. a b
26.
revenue = 3X + 10Y, so E[revenue] = E[3X + 10Y] = x:05y:02(3x+10y) * p(x,y) = 0 * p(0,0) + ...+35 * p(5,2) = 15.4
29.
Cov(X,Y) = -2/75 and x = y = (2/5);E(X2) = 01x2*fx(x)dx = 1201x3(1-x)2dx = 12/60 = 1/5, so Var(X) = (1/5) - (4/25) = 1/25 Similarly, Var(Y) = 1/25, so x,y = (-2/75)/[(1/25)0.5 * (1/25)0.5] = -50/75 = -0.667
7.
Clark Data Set
Probability Plot of weights
Normal 70 65 60 weights 55 50 45 40 -3 -2 -1 0 Score 1 2 3
Mean 55.66 StDev 3.905 N 220 AD 4.377 P-Value <0.005
Looking at the Normal Probability plot it is clear that the data in the plot is not really along a straight line the Normality assumption would be inappropriate here. (However, it does appear that things might be straight in the middle so perhaps if we only looked at data between (say) 53 and 60 the Normality assumption might be more appropriate however, this would assume that all of the extreme values are outliers, and to be certain one would need to drop the data outside this range and redo the plot....)