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HW8 - 6.13 z = —1.88 I = URN—1.88 10 = 6.24 years...

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Unformatted text preview: 6.13 z = —1.88, I = URN—1.88] + 10 = 6.24 years. 6.14 (a) z = (159.75 — 1743/63 = —2.14; P(X < 159.75} = P(Z c: —2.14) = 0.0162. Therefore, [1000){06162} = 16 students. (5) z. = (171.25 — 174.5)1’59 = —0.47, 22 = (152.25 — 174.5)[59 = 1.12. P(171.25 < X 4 152.25) = P(—0.47 < Z < 1.12) = 0.5555 — 0.5192 = 0.5494. Therefore, {1060){05494} = 549' students. (:3) z, = (174.75 — 174.5);09 = 0.04. z; = (175.25 — 174.5};‘09 = 0.11. P(174.75 < X < 175.25] = P(U.04 < Z < {L11} = 0.5433 — 0.5150 = 0.0278. Therefore, {1060)(U.0278)=23 students. (d) z = (157.75 — 17451159 = 1.92, PiX 2 157.75) = P(Z > 1.92) = 0.0274. Therefore, {1060)(0fl274) = 27 students. 6.21 A=TandB=IU. (5 } (PX- <1 5. 5): =0. 50. (5 1 P(7.4 4 X 4 951—922“ _ 0.70. (c) P(X 2 5.5) = 1033-5 = 0.50. 5.22 (a) P(X 2- 7)_ — “3,57 = 0.5. (h) P(2<X<7}=”+,=0.5. 6.23 {a} From Table 41.1 with n. = 15 and p = [1.2 we have 4 PU 3' X i: 4} = Z b[2,15,0.2)— bffl,15,0.2) = 0.8358 — [1.6352 = 0.8606. 3:0 (b) By the normal—curve approximation we first find 11 = 1121 = 3 and then 0‘2 = mpg = {15}(0.2]{0.3} = 2.4. Then 0' = 1.549. Now, 21 = (0.5— 3),!1549 = —1.61 and .52 = (4.5 — 3],“1549 = [1.91 Therefore, P(1 51 X g: 4) = P{—1.61 51 Z 5 0.97) = [1.3341] — 6.0537 2 0.7863. 6.24 ,1: = 21;) = {400){1f2} = 200, o = #2121 = (4OU)(1,’2)(1/2) = (a}z={134.5—200U10= —1.55 and 22: (210—5 200)/10=1..05 (191.84 5 c: X < 210 5): P(— 1. 55 < Z < 1. 0521— — 0. 5531 — 0.0000— — 0.7025. (5} z. = {204.5 — 2001,20 = 0.45 and 25 = (205.5 — 2001210 = 0.55. PRU-4.5 < X 4 205.5) = P(0.45 < z < 0.55} = 0.7055 — 0.5255 = 0.0552. (c) 21 = {175.5 — 260)]10 = —2.45 and .52 = (227.5 — 200)}10 = 2.75. PO: < 175.5) + P(X > 227.5) = P{Z < —2.45) + P{Z > 2.75} = P(Z < —2.45) + 1 — P(Z < 2.75) = 0.00?1+ 1 — 0.9070 = 0.0101. 5.33 11 = (400)11/10) = 40 and 5 = ((400) 1/10) {)_9/10 — (5 ) z— (31. 5— 401/5— _ —1.42; P(X 4 31. 5): P(Z < —1.42) = 0.0775. (5): =(49.5 — 401/5 = 1.55: 13‘an > 49.5) = P(Z > 1.55) = 1 — 0.9429 = 0.0571. 55) ={34 5— 4U}/6— — —U. 92 and 22: {46.5 — 40)}6 = 1.08; 10(34 5 < X «4 45 5): P(— —.0 92 < z < 1.05) = 0.5599 — 0.1755 = 0.5511. 6.39 P(1.3 < X < 2.4) = ff: re': (£3 = [—re'z — 3-1“ij = see-L3 — 3.4.94”1 = 0.1545. 6.45 P(X < 3) = i]: 3‘3“ air = —8_I'ME: = 1 — e‘af“ = 0.5276. Let Y be the number of days a person is served in less than 3 minutes. Then 6 PW 2 4) = Z b{y;6.1 — 3-”) = (3) (0.5276)‘(n.4724)2 + {2)(n.5276)5(0.4724) 1:4 + {2) (0.52m)fl = 0.3968. 5.46 P{X < 1} = if; 3‘3” d3 = —e‘T”E; = 1— 8—1” = 0.3935. Let Y be the number of switches that fail during the first year. Using the normal approximation we find ,u = {lUU}{U.3935} = 39.35, 0' = 3/{100}{U.3935}{U.6065} = 4.885, and z = {30.5 — 39.35)f4.885 = —1.81. Therefore, PU” <_-’. 30} = P{Z < —l.81} = [13352. 8.4 {a} :E = 8.6 minutes. (b) :E = 9.5 minutes. (:3) Mode are 5 and 10 minutes. 8.9 {a} Range = 15 — 5 = ID. ”I:EEIEE_EE::1I*J2 {b} s2 = W = W = 10.933. Taking the square root, we have s = 3.337. nifi4ih? 3.13 32 = fl = W = [1.342 and hence s = 0.585. 3.14 [a] Replace X.- in 5'2 by X.- + c for i = 1,2,. . .,n. Then 32' becomes X +c and fl Ewe“) — one]? = 92:]. l 1 20:, — X}? n—1_ (b) Replace X,- by cX, in 32 for 1' = 1,2, .. .,n. Then 1? becomes a)? and S2 = 1 1202):, —c.r}? = :1 Ear, 4?)? i=1 3:1 ...
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