SOLUTIONS FOR THE STATISTICS 3083 FINAL OF DECEMBER 1996
 QUESTION 1 Arthritis Arthritis Total Free P[Arthritis | Positive] Positive 0.12 0.034 0.154 0.12 / 0.154 Negative 0.03 0.816 0.846 Total 0.15 0.850 1.000 QUESTION 2 t 1 2 3 4 5 SUM f(t) 0.15 0.18 0.27 0.22 0.18 1 F(t) 0.15 0.33 0.6 0.82 1 ... Cumulative f(t)*t 0.15 0.36 0.81 0.88 0.9 3.1 = E(T) D = t-3.1 -2.1 -1.1 -0.1 0.9 1.9 ... f(t)*D*D 0.6615 0.2178 0.0027 0.1782 0.6498 1.71 = V(T) (a) E(T)=3.1 V(T)=1.71 (b) (0.27+0.22+0.18) / (0.18+0.27+0.22+0.18) = 67 / 85 (c) (2*0.18+3*0.27+4*0.22+5*0.18) / (0.18+0.27+0.22+0.18) = 277 / 85 (d) See the "Cumulative" row of the calculations QUESTION 3 Presumably we are to assume (i) The 50% rate applies every night. (ii) Customers are statistically independent. -- This may not, in fact, be true, for example, will there -- be a heavier demand for sets on nights of the Grey Cup? S = # of sets 0 1 2 3 4 5 SUM Probability 1/32 5/32 10/32 10/32 5/32 1/32 1 Rental fee 0 2 4 6 6 6 Prob*Rental 0 10/32 40/32 60/32 30/32 6/32 146/32 (a) P[S > 3 ] 5/32 + 1/32 = 6 / 32 (evenings) (b) E( Rental ) 146/32 = 4.5625 (dollars) QUESTION 4 (a) c = 1/2 (b) F(y) = 0 for y < 0 = 1 for u > 1 = y - (y*y/4) otherwise (c) F(1)/F(1.5) = 0.75/0.9375 = 0.8 (d) This is integral from 0 to 1.5 of (0.5)(2-y) y dy = 0.5625 divided by F(1.5) = 0.9375 to get the quotient 0.6 (e) Hazard rate: f(y) / [1-F(y)] (1-y/2) / [1 - y + y*y/4] = 1 /[1 - y/2] QUESTION 5 (a) The cumulative function is F(X) = 1-exp(-X/11) To find the half life, solve F(X) = 1/2 X = 11 ln(2) (b) The exponential distribution having no memory, this is the 6 plus the unconditional expected life, i.e. 6 + 11 = 17 (c) Using again the memoryless property, this equals the proba- bility that a new component lasts 5 hours or more which is 1 - F(5) = exp(-5/11) QUESTION 6 (a) z = (100-120)/30 = -2/3 P[Z > -2/3] = 0.252 (b) Inversely look up 0.57 in a table of the normal distribution to get z = 0.18 QUESTION 7 The hazard function is the derivative of the negative (natural) log complimentary cumulative function, 1-F. Integrate 6t to get the the negative logarithm of 1-F(t) -ln (1-F(t)) = 3t2 + constant 1-F(t) = [another const] exp(-3t2) The constant is chosen to make 1-F(0)=1 and thus is one. f(t) = 6 t exp(-3t2) This is a Weibull distribution, with alpha = 3, beta = 2 . Expectation and variance can be found in textbook table 4.8 E{t} = sqrt(1/3) G(3/2) V{t} = (1/3)G(2) - (E{t})2 Note: G is used in place of upper case Gamma which is not available in HTML. QUESTION 8 (a) X and Y are not independent. On inspection the body of the table is seen not to be the product of its marginal sums. (b) The sum of these numbers: . . . . . . 0.12 0.05 . . 0.05 0.00 . . 0.08 0.00 which is 0.3 (c) E(X) = 4.1 E(Y) = 2.3 E(XY)= 9.24 Cov = 9.24 - 4.1 2.3 = -0.19 (d) (5.76+3.2+5.4+0+15.36+0)/(0.12+0.05+0.05+0+0.08+0) = 29.72 / 0.3 (e) The sum of these numbers: 0.12 0.08 0.15 . which is 0.42 0.07 . . . . . . . . . . . QUESTION 9 Use the central limit theorem to approximate a probability: E{ W } = 40*36 + 40*30 = 2640 V{ W } = 40*25 + 40*25 = 2000 std = 44.72 z = (2600-2640) / 44.72 = -0.894 Probability = 0.186 QUESTION 10 W is chi square with 90 degrees of freedom. Near the middle the chi square distrtibution with many degrees of freedom is well approximated by a normal distribution with the same mean and variance. E{W} = 90 V{W} = 180 (a) The normal distribution being symmetric, the probability is 1/2. (b) std{W} = 13.42 z = (100-90)/ 13.42 = -0.745 Probability = 0.77