Semester Two Final Examinations, 2019 STAT2203 Probability Models and Data Analysis for Engineering Page 1 of 18 This exam paper must not be removed from the venue School of Mathematics & Physics EXAMINATION Semester Two Final Examinations, 2019 STAT2203 Probability Models and Data Analysis for Engineering This paper is for St Lucia Campus students. Examination Duration: 120 minutes Reading Time: 10 minutes Exam Conditions: This is a Central Examination This is a Closed Book Examination - specified materials permitted During reading time - write only on the rough paper provided This examination paper will be released to the Library Materials Permitted In The Exam Venue: (No electronic aids are permitted e.g. laptops, phones) Calculators - Casio FX82 series or UQ approved (labelled) One A4 sheet of handwritten notes double sided is permitted Materials To Be Supplied To Students: None Instructions To Students: Additional exam materials (eg. rough paper) will be provided upon request. There are 50 marks available on this exam from 5 questions. Write your answers in the spaces provided on pages 2 – 13 of this examination paper. Show your working and state conclusions where appropriate. Pages 14 – 18 give formulas and statistical tables. Those pages will not be marked. Venue ____________________ Seat Number ________ Student Number |__|__|__|__|__|__|__|__| Family Name _____________________ First Name _____________________ For Examiner Use Only Question Mark Total ________ STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 1. [10 marks] A pair of random variables (X, Y ) has a joint probability distribution in which X ∼ Uniform(0, 1), and (Y |X = x) ∼ Uniform(−x, 2x); that is, the conditional distribution of Y given {X = x} is uniform on the interval (−x, 2x). (a) Write down the joint probability density function of (X, Y ), clearly specifying the support of the distribution. [1 mark] (b) Determine the marginal probability density function of Y . [3 marks] Page 2 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (c) Using the formula EY = E [E [Y |X]], find the expectation of Y . [3 marks] (d) Compute Cov(X, Y ). [3 marks] Page 3 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 2. [6 marks] Let X be the continuous random variable with probability density function fX(x) = { − log(x), x ∈ (0, 1) 0, otherwise. Define the random variable Y = − log(X). (a) Find the probability density function of Y . [3 marks] (b) The moment generating function of Y is MY (t) = E(etY ) = (1− t)−2, t < 1. Determine the expected value and variance of Y . [3 marks] Page 4 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 3. [10 marks] A study compared the short term effect of playing computer games and watching television on young children’s aggression. In the study 28 children aged 4 years to 6 years were randomly assigned to one of two groups with 14 children being assigned to each group. The children were monitored before and after these activities and the level of aggressive behaviour rated on a scale from 1 (no aggressive behaviour) to 100 (very aggressive behaviour). (a) The first group of children, those who watched television, experienced an aver- age increase of 5.88 in their aggressive behaviour rating with a sample standard deviation of 6.44. Construct a 95% confidence interval for the mean change in aggressive behaviour rating for children after watching television. What does this confidence interval imply in terms of the mean change in aggressive be- haviour rating? [4 marks] (b) If the number of children in the study was doubled, but all other sample statistics remained the same, what would happen to the width of the confidence interval? [1 mark] Page 5 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (c) The children in the second group, those who played a computer game, expe- rienced an average increase of 3.50 in their aggressive behaviour rating with a sample standard deviation of 8.00. Is there any evidence of a difference in the mean change in aggression rating between the two groups? State the null and alternative hypotheses, and use an appropriate test statistic to determine the P -value. What do you conclude? [5 marks] Page 6 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 4. [10 marks] A study was conducted in Germany on the demographics of the online gaming population. Part of this study involved interviewing 688 adult volunteers about their online gaming experience. (a) The volunteers were asked what type of games they played online (Role-playing, Action, Strategy, Sports and Racing). Of the 464 men interviewed, 325 said they play role-playing games while of the 224 women interviewed 178 said they play role-playing games. Is there evidence of a difference in the proportion of men and women who play role playing games online? State the null and alternative hypotheses, and use an appropriate test statistic to determine the P -value. What do you conclude? [4 marks] Page 7 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (b) The volunteers were also asked how many years have they been playing online games. The responses are summarised in the table below. Based on this table, is there evidence of an association between sex and years of experience in online gaming? [6 marks] Page 8 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 5. [14 marks] Twenty seven right-handed university students were recruited into a study to investigate the time taken for a person to move a cursor on a computer screen along a specified path of length A and width W using a mouse. Each student was given a single path along which they were to move the cursor and the time (in ms) taken to complete the task was recorded. The data are displayed in the figure below together with the fitted least squares line. The output on the next page shows the results of a linear regression fit in MATLAB for the relationship between the time taken to complete the task (MovementTime) and the ratio A/W (AWratio). Page 9 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (a) Briefly interpret the value 46.299 in the regression output. [1 mark] (b) Briefly interpret the value 211 in the regression output. [1 mark] Page 10 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (c) Give a 95% confidence interval for the underlying slope of the linear relationship between the time taken to complete the task and the ratio A/W . [2 marks] (d) Drury’s law suggests that the time taken to complete the task will depend on the ratio A/W in the following way: Movement Time = Constant× (A/W ) . Does the data provide evidence that when the ratio A/W is zero, the time taken to complete the task will also be zero? State the null and alternative hypotheses, and report the appropriate test statistic and P -value from the output. What do you conclude? [3 marks] Page 11 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (e) The following figures were generated in MATLAB to help check the assumptions underlying the linear regression. Comment on the validity of the assumptions underlying linear regression for this data with reference to these figures and the figure on page 9. [3 marks] Page 12 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 (additional space for answer to part (e)) (f) The covariance matrix for the estimator of (intercept, slope) is[ 14591 −628 −628 30 ] . Construct a 95% confidence interval for the expected time taken to complete the task when the ratio A/W is 20. [4 marks] END OF EXAMINATION Page 13 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 Formula Sheet Elementary probability • Sum rule: For disjoint A1, A2, . . .: P( ⋃ i Ai) = ∑ i P(Ai). • P(Ac) = 1− P(A). • P(A ∪B) = P(A) + P(B)− P(A ∩B). • Conditional probability: P(A |B) = P(A∩B)P(B) . • Law of total probability: P(A) = ∑n i=1 P(A |Bi)P(Bi), where B1, B2, . . . , Bn is a partition of Ω. • Bayes’ Rule: P(Bj |A) = P(Bj) P(A|Bj)∑n i=1 P(Bi) P(A|Bi) . • Independent events: P(A ∩B) = P(A)P(B) . Random variables • Cdf of X: F (x) = P(X 6 x), x ∈ R. • Pmf of X: (discrete r.v.) f(x) = P(X = x). • Pdf of X: (continuous r.v.) f(x) = F ′(x). • For a discrete r.v. X: P(X ∈ B) = ∑ x∈B P(X = x). • For a continuous r.v. X with pdf f : P(X ∈ B) = ∫ B f(x) dx. • In particular (continuous), F (x) = ∫ x−∞ f(u) du. • Important discrete distributions: Distr. pmf suppport Ber(p) px(1− p)1−x {0, 1} Bin(n, p) (n x ) px(1− p)n−x {0, 1, . . . , n} Poi(λ) e−λ λ x x! {0, 1, . . .} Geom(p) p(1− p)x−1 {1, 2, . . .} • Important continuous distributions: Distr. pdf x ∈ U[a, b] 1 b−a [a, b] Exp(λ) λ e−λx R+ N(µ, σ2) 1 σ √ 2pi e − 1 2 ( x−µ σ )2 R • Expectation (discr.): EX = ∑x xP(X = x). • (of function) E g(X) = ∑x g(x)P(X = x) . • Expectation (cont.): EX = ∫ x f(x) dx. • (of function) E g(X) = ∫ g(x)f(x) dx, • EX and Var(X) for discrete distributions: EX Var(X) Ber(p) p p(1− p) Bin(n, p) np np(1− p) Geom(p) 1 p 1−p p2 Poi(λ) λ λ • EX and Var(X) for continuous distributions: EX Var(X) U(a, b) a+b 2 (b−a)2 12 Exp(λ) 1 λ 1 λ2 N(µ, σ2) µ σ2 Multiple random variables • Joint distribution: P((X,Y ) ∈ B) = ∫∫B fX,Y (x, y) dx dy . • Marginal pdf: fX(x) = ∫ fX,Y (x, y) dy. • Independent r.v.’s: fX1,...,Xn (x1, . . . , xn) = ∏n k=1 fXk (xk) . • Expected sum : E(aX + bY ) = aEX + bEY . • Expected product (if X,Y independent): E[X Y ] = EX EY. • Markov inequality: P(X > x) 6 EX x . • Covariance: cov(X,Y ) = E(X − EX)(Y − EY ). • Properties of Var and Cov: Var(X) = EX2 − (EX)2. Var(aX + b) = a2Var(X). cov(X,Y ) = EXY − EXEY . cov(X,Y ) = cov(Y,X). cov(aX + bY, Z) = a cov(X,Z) + b cov(Y, Z). cov(X,X) = Var(X). Var(X + Y ) = Var(X) + Var(Y ) + 2 cov(X,Y ). X and Y independent =⇒ cov(X,Y ) = 0. • Conditional pdf: If fX(x) > 0, fY |X(y |x) := fX,Y (x,y)fX (x) , y ∈ R. • The corresponding conditional expectation: E[Y |X = x] = ∫ y fY |X(y |x)dy. • E[Y ] = E[E[Y |X]] • Moment Generating Function (MGF): When it exists, for t ∈ I ⊂ R, M(t) = E etX = ∫∞ −∞ e tx f(x) dx . Page 14 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 • MGFs for various distributions: U(a, b) e bt−eat t(b−a) Exp(λ) ( λ λ−s ) N(µ, σ2) etµ+σ 2t2/2 • Moment property: EXn = M(n)(0). • MX+Y (t) = MX(t)MY (t), ∀t, if X,Y independent. • If Xi ∼ N(µi, σ2i ) (independent), then a+ ∑n i=1 biXi ∼ N ( a+ ∑n i=1 bi µi, ∑n i=1 b 2 i σ 2 i ) . • Pdf of the multivariate Normal distribution: fZ(z) = 1√ (2pi)n |Σ| e − 1 2 (z−µ)TΣ−1(z−µ) . Σ is the covariance matrix, and µ the mean vector. • If X has a multivariate Normal distribution N(µ,Σ) (dimension n) and Y = a+BX (dimension m 6 n), then Y ∼ N(a+Bµ, BTΣB). • Central Limit Theorem: lim n→∞P ( Sn − nµ σ √ n 6 x ) = Φ(x), where Φ is the cdf of the standard Normal distribu- tion. • Normal Approximation to Binomial: If X ∼ Bin(n, p), then, for large n, P(X 6 k) ≈ P(Y 6 k), where Y ∼ N(np, np(1− p)) . Statistics Tests and Confidence Intervals Based on Standard Errors • Test statistic: estimate−hypothesised se(estimate) . • Confidence interval: estimate ± (critical value)× se(estimate). • se(x¯) = s√ n • se(x¯− y¯) = sp √ 1 nx + 1 ny • (pooled sample variance) s2p = (nx−1)s2x+(ny−1)s2y nx+ny−2 . • se(p̂) = √ p̂(1−p̂) n • se(p̂x − p̂y) = √ p̂x(1−p̂x) nx + p̂y(1−p̂y) ny • Use t-distribution for means, correlation and regres- sion. Use normal distribution for proportions. Chi-squared test • expected count = (row total)×(column total) overall total . • X2 = ∑ (observed−expected)2 expected • degrees of freedom = (#rows− 1)× (#columns− 1). Linear regression • Y ∼ N(Xβ, σ2I) • estimator β̂ = (XTX)−1XTY • Cov(β̂) = σ2(XTX)−1 • s2 = (Y−Xβ̂)T (Y−Xβ̂) n−p Other Mathematical Formulas • Factorial. n! = n (n− 1) (n− 2) · · · 1. Gives the num- ber of permutations (orderings) of {1, . . . , n}. • Binomial coefficient. (n k ) = n! k! (n−k)! . Gives the num- ber combinations (no order) of k different numbers from {1, . . . , n}. • Newton’s binomial theorem: (a + b)n =∑n k=0 a k bn−k. • Geometric sum: 1 + a + a2 + · · · + an = 1−an+1 1−a (a 6= 1). If |a| < 1 then 1 + a+ a2 + · · · = 1 1−a . • Logarithms: 1. log(x y) = log x+ log y. 2. elog x = x. • Exponential: 1. ex = 1 + x+ x 2 2! + x 3 3! + · · · . 2. ex = limn→∞ ( 1 + x n )n . 3. ex+y = ex ey . • Differentiation: 1. (f + g)′ = f ′ + g′ 2. (fg)′ = f ′g + fg′ 3. ( f g )′ = f ′g−fg′ g2 4. d dx xn = nxn−1 5. d dx ex = ex 6. d dx log(x) = 1 x • Chain rule: (f(g(x)))′ = f ′(g(x)) g′(x) . • Integration: ∫ ba f(x) dx = [F (x)]ba = F (b) − F (a), where F ′ = f . • Integration by parts: ∫ ba f(x)G(x) dx = [F (x)G(x)]ba − ∫ b a F (x) g(x) dx . (Here F ′ = f and G′ = f .) Page 15 of 18 Table 12.1: Standard Normal distribution Second decimal place of z z 0 1 2 3 4 5 6 7 8 9 0.0 0.500 0.496 0.492 0.488 0.484 0.480 0.476 0.472 0.468 0.464 0.1 0.460 0.456 0.452 0.448 0.444 0.440 0.436 0.433 0.429 0.425 0.2 0.421 0.417 0.413 0.409 0.405 0.401 0.397 0.394 0.390 0.386 0.3 0.382 0.378 0.374 0.371 0.367 0.363 0.359 0.356 0.352 0.348 0.4 0.345 0.341 0.337 0.334 0.330 0.326 0.323 0.319 0.316 0.312 0.5 0.309 0.305 0.302 0.298 0.295 0.291 0.288 0.284 0.281 0.278 0.6 0.274 0.271 0.268 0.264 0.261 0.258 0.255 0.251 0.248 0.245 0.7 0.242 0.239 0.236 0.233 0.230 0.227 0.224 0.221 0.218 0.215 0.8 0.212 0.209 0.206 0.203 0.200 0.198 0.195 0.192 0.189 0.187 0.9 0.184 0.181 0.179 0.176 0.174 0.171 0.169 0.166 0.164 0.161 1.0 0.159 0.156 0.154 0.152 0.149 0.147 0.145 0.142 0.140 0.138 1.1 0.136 0.133 0.131 0.129 0.127 0.125 0.123 0.121 0.119 0.117 1.2 0.115 0.113 0.111 0.109 0.107 0.106 0.104 0.102 0.100 0.099 1.3 0.097 0.095 0.093 0.092 0.090 0.089 0.087 0.085 0.084 0.082 1.4 0.081 0.079 0.078 0.076 0.075 0.074 0.072 0.071 0.069 0.068 1.5 0.067 0.066 0.064 0.063 0.062 0.061 0.059 0.058 0.057 0.056 1.6 0.055 0.054 0.053 0.052 0.051 0.049 0.048 0.047 0.046 0.046 1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.037 1.8 0.036 0.035 0.034 0.034 0.033 0.032 0.031 0.031 0.030 0.029 1.9 0.029 0.028 0.027 0.027 0.026 0.026 0.025 0.024 0.024 0.023 2.0 0.023 0.022 0.022 0.021 0.021 0.020 0.020 0.019 0.019 0.018 2.1 0.018 0.017 0.017 0.017 0.016 0.016 0.015 0.015 0.015 0.014 2.2 0.014 0.014 0.013 0.013 0.013 0.012 0.012 0.012 0.011 0.011 2.3 0.011 0.010 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.008 2.4 0.008 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 2.5 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.005 0.005 2.6 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.004 0.004 2.7 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 2.8 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 2.9 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 3.0 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.1 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.2 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.3 This table gives P (Z z ) for Z ⇠ Normal(0,1). Critical values of the Normal distribution, the z ⇤ values such that P (Z z ⇤) = p for a particular p , can be found from the 1 row of Table 14.2. Page 16 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 Table 14.2: Critical values of Student’s T distribution Probability p df 0.25 0.10 0.05 0.025 0.01 0.005 0.001 0.0005 0.0001 1 1.000 3.078 6.314 12.71 31.82 63.66 318.3 636.6 3183.1 2 0.816 1.886 2.920 4.303 6.965 9.925 22.33 31.60 70.70 3 0.765 1.638 2.353 3.182 4.541 5.841 10.21 12.92 22.20 4 0.741 1.533 2.132 2.776 3.747 4.604 7.173 8.610 13.03 5 0.727 1.476 2.015 2.571 3.365 4.032 5.893 6.869 9.678 6 0.718 1.440 1.943 2.447 3.143 3.707 5.208 5.959 8.025 7 0.711 1.415 1.895 2.365 2.998 3.499 4.785 5.408 7.063 8 0.706 1.397 1.860 2.306 2.896 3.355 4.501 5.041 6.442 9 0.703 1.383 1.833 2.262 2.821 3.250 4.297 4.781 6.010 10 0.700 1.372 1.812 2.228 2.764 3.169 4.144 4.587 5.694 11 0.697 1.363 1.796 2.201 2.718 3.106 4.025 4.437 5.453 12 0.695 1.356 1.782 2.179 2.681 3.055 3.930 4.318 5.263 13 0.694 1.350 1.771 2.160 2.650 3.012 3.852 4.221 5.111 14 0.692 1.345 1.761 2.145 2.624 2.977 3.787 4.140 4.985 15 0.691 1.341 1.753 2.131 2.602 2.947 3.733 4.073 4.880 16 0.690 1.337 1.746 2.120 2.583 2.921 3.686 4.015 4.791 17 0.689 1.333 1.740 2.110 2.567 2.898 3.646 3.965 4.714 18 0.688 1.330 1.734 2.101 2.552 2.878 3.610 3.922 4.648 19 0.688 1.328 1.729 2.093 2.539 2.861 3.579 3.883 4.590 20 0.687 1.325 1.725 2.086 2.528 2.845 3.552 3.850 4.539 21 0.686 1.323 1.721 2.080 2.518 2.831 3.527 3.819 4.493 22 0.686 1.321 1.717 2.074 2.508 2.819 3.505 3.792 4.452 23 0.685 1.319 1.714 2.069 2.500 2.807 3.485 3.768 4.415 24 0.685 1.318 1.711 2.064 2.492 2.797 3.467 3.745 4.382 25 0.684 1.316 1.708 2.060 2.485 2.787 3.450 3.725 4.352 26 0.684 1.315 1.706 2.056 2.479 2.779 3.435 3.707 4.324 27 0.684 1.314 1.703 2.052 2.473 2.771 3.421 3.690 4.299 28 0.683 1.313 1.701 2.048 2.467 2.763 3.408 3.674 4.275 29 0.683 1.311 1.699 2.045 2.462 2.756 3.396 3.659 4.254 30 0.683 1.310 1.697 2.042 2.457 2.750 3.385 3.646 4.234 40 0.681 1.303 1.684 2.021 2.423 2.704 3.307 3.551 4.094 50 0.679 1.299 1.676 2.009 2.403 2.678 3.261 3.496 4.014 60 0.679 1.296 1.671 2.000 2.390 2.660 3.232 3.460 3.962 70 0.678 1.294 1.667 1.994 2.381 2.648 3.211 3.435 3.926 80 0.678 1.292 1.664 1.990 2.374 2.639 3.195 3.416 3.899 90 0.677 1.291 1.662 1.987 2.368 2.632 3.183 3.402 3.878 100 0.677 1.290 1.660 1.984 2.364 2.626 3.174 3.390 3.862 1 0.674 1.282 1.645 1.960 2.326 2.576 3.090 3.291 3.719 This table gives t ⇤ such that P (T t ⇤) = p , where T ⇠Student(df). Page 17 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019 Table 22.4: 2 distribution Probability p df 0.975 0.95 0.25 0.10 0.05 0.025 0.01 0.005 0.001 1 0.001 0.004 1.323 2.706 3.841 5.024 6.635 7.879 10.83 2 0.051 0.103 2.773 4.605 5.991 7.378 9.210 10.60 13.82 3 0.216 0.352 4.108 6.251 7.815 9.348 11.34 12.84 16.27 4 0.484 0.711 5.385 7.779 9.488 11.14 13.28 14.86 18.47 5 0.831 1.145 6.626 9.236 11.07 12.83 15.09 16.75 20.52 6 1.237 1.635 7.841 10.64 12.59 14.45 16.81 18.55 22.46 7 1.690 2.167 9.037 12.02 14.07 16.01 18.48 20.28 24.32 8 2.180 2.733 10.22 13.36 15.51 17.53 20.09 21.95 26.12 9 2.700 3.325 11.39 14.68 16.92 19.02 21.67 23.59 27.88 10 3.247 3.940 12.55 15.99 18.31 20.48 23.21 25.19 29.59 11 3.816 4.575 13.70 17.28 19.68 21.92 24.72 26.76 31.26 12 4.404 5.226 14.85 18.55 21.03 23.34 26.22 28.30 32.91 13 5.009 5.892 15.98 19.81 22.36 24.74 27.69 29.82 34.53 14 5.629 6.571 17.12 21.06 23.68 26.12 29.14 31.32 36.12 15 6.262 7.261 18.25 22.31 25.00 27.49 30.58 32.80 37.70 16 6.908 7.962 19.37 23.54 26.30 28.85 32.00 34.27 39.25 17 7.564 8.672 20.49 24.77 27.59 30.19 33.41 35.72 40.79 18 8.231 9.390 21.60 25.99 28.87 31.53 34.81 37.16 42.31 19 8.907 10.12 22.72 27.20 30.14 32.85 36.19 38.58 43.82 20 9.591 10.85 23.83 28.41 31.41 34.17 37.57 40.00 45.31 21 10.28 11.59 24.93 29.62 32.67 35.48 38.93 41.40 46.80 22 10.98 12.34 26.04 30.81 33.92 36.78 40.29 42.80 48.27 23 11.69 13.09 27.14 32.01 35.17 38.08 41.64 44.18 49.73 24 12.40 13.85 28.24 33.20 36.42 39.36 42.98 45.56 51.18 25 13.12 14.61 29.34 34.38 37.65 40.65 44.31 46.93 52.62 26 13.84 15.38 30.43 35.56 38.89 41.92 45.64 48.29 54.05 27 14.57 16.15 31.53 36.74 40.11 43.19 46.96 49.64 55.48 28 15.31 16.93 32.62 37.92 41.34 44.46 48.28 50.99 56.89 29 16.05 17.71 33.71 39.09 42.56 45.72 49.59 52.34 58.30 30 16.79 18.49 34.80 40.26 43.77 46.98 50.89 53.67 59.70 40 24.43 26.51 45.62 51.81 55.76 59.34 63.69 66.77 73.40 50 32.36 34.76 56.33 63.17 67.50 71.42 76.15 79.49 86.66 60 40.48 43.19 66.98 74.40 79.08 83.30 88.38 91.95 99.61 70 48.76 51.74 77.58 85.53 90.53 95.02 100.4 104.2 112.3 80 57.15 60.39 88.13 96.58 101.9 106.6 112.3 116.3 124.8 90 65.65 69.13 98.65 107.6 113.1 118.1 124.1 128.3 137.2 100 74.22 77.93 109.1 118.5 124.3 129.6 135.8 140.2 149.4 This table gives x ⇤ such that P (X 2 x ⇤) = p , where X 2 ⇠ 2(df). Page 18 of 18 STAT2203: Probability Models and Data Analysis for Engineering Final Exam, Semester 2, 2019
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