Chapter 5
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Chapter 5 Supplementary Exercises 60. a. fx) 0.150 0.050 0.075 0.050 0.125 0.050 0.100 0.125 X 1 2 3 4 5 6 7 8 9 0.125 10 0.150 Total 1.000 b. Probability of outstanding service is 0.125 + 0.150 = 0.275 c X fx) xf(x) X—p (x = p)? (x = p)? Ax) 1 0.150 0.150 -4.925 24.2556 3.6383 2 0.050 0.100 -3.925 15.4056 0.7703 3 0.075 0.225 -2.925 8.5556 0.6417 4 0.050 0.200 -1.925 3.7056 0.1853 5 0.125 0.625 -0.925 0.8556 0.1070 6 0.050 0.300 0.075 0.0056 0.0003 7 0.100 0.700 1.075 1.1556 0.1156 8 0.125 1.000 2.075 4.3056 0.5382 9 0.125 1.125 3.075 9.4556 1.1820 10 0.150 1.500 4.075 16.6056 2.4908 Total 1.000 5.925 9.6694 KXx) = 5.925 and Var(x) = 9.6694 d. The probability of a new car dealership receiving an outstanding wait-time rating is 2/7 = 0.2857. For the remaining 40 — 7 = 33 service providers, nine received an outstanding rating; this corresponds to a probability of 9/33 = 0.2727. For these results, there does not appear to be much difference between the probability that a new car dealership is rated outstanding compared to the same probability for other types of service providers.
62. a. There are 600 observations involving the two variables. Dividing the entries in the table shown by 600 and summing the rows and columns we obtain the following. Reading Material (y) Snacks (x) 0 1 2 Total 0 0.0 0.1 0.03 0.13 1 0.4 0.15 0.05 0.6 2 0.2 0.05 0.02 0.27 Total 0.6 0.3 0.1 1 The entries in the body of the table are the bivariate or joint probabilities for xand y. The entries in the right most (Total) column are the marginal probabilities for x, and the entries in the bottom (Total) row are the marginal probabilities for y. The probability of a customer purchasing one item of reading materials and two snack items is given by { x=2, y=1) = 0.05. The probability of a customer purchasing one snack item only is given by {x = 1, y = 0) = 0.40. The probability £x = 0, y = 0) = 0 because the point of sale terminal is only used when someone makes a purchase. 60 b. The marginal probability distribution of x along with the calculation of the expected value and variance is shown as follows. x fx) ) x-Kx (x- (x- §x1)* £x)* 1) 0 0.13 0 -1.14 1299 0.1689 1 0.60 0.6 -0.14 0.0196 0.0118 2 027 0.54 0.86 0.7396 0.1997 114 0.3804 =Var(x) =£x) We see that £(x) = 1.14 snack items and Var(x) = 0.3804. c. The marginal probability distribution of y along with the calculation of the expected value and variance is shown as follows. y L) my) y- 8y (y- (y- BEN) A»)? L¢7) 0 0.60 0 -05 0.25 0.15 1 0.30 03 0.5 0.25 0.075 2 0.10 0.2 15 2.25 0.225 05 0.45 =E(y) =Var(y) We see that £(y) = 0.50 reading materials and Var(y) = 0.45.
d. The probability distribution of = x + yis shown as follows along with the calculation of its expected value and variance. t LO) ne) T-Rf) (¢-KRp)* (- AH) 19 1 0.50 -0.64 0.409% 0.2048 2 038 0.36 0.129% 0.0492 3 0.10 136 1.8496 0.1850 4 0.02 2.36 5.5696 0.1114 1.64 0.5504 =E1) =Var(t) We see that the expected number of items purchased is £(¢) = 1.64, and the variance in the number of purchases is Var(t) = 0.5504. e. From part (b), Var(x) = 0.3804. From part (c), Var(y) = 0.45. And from part (d), Var(x + y) = Var(t) = 0.5504. Therefore, o, =WVar(x+y)~Var(x)~Var(y)]/ 2 =(0.5504-0.3804-0.4500)/ 2 =-0.14 To compute the correlation coefficient, we must first obtain the standard deviation of xand y. o, = \/Vnr(x) =/0.3804 = 0.6168 o, = Var(y) =0.45 = 0.6708 61 So, the correlation coefficient is given by Py == =03384 The relationship between the number of reading materials purchased and the number of snacks purchased is negative. This means that the more reading materials purchased the fewer snack items purchased and vice versa.
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