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# stat400sln6 - 4.2—2 p 03 U2 2 1/3 ﬂx —1.o"-o.s...

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Unformatted text preview: 4.2—2 p, : 03 U2 2 1/3. ﬂx) —1.o "-o.s * —5.2 0.2 0.6 1.0 4.2-8 (a) ﬁx) 2 (fut/3, 0 S :1: < oo; 002— . —: w__ —43 (b)P(X>2):/2 3e 2Z/3dx:[—e 2/3]2 ——e /. m 4.2.18. (3.) P(X>40) : / ﬁ8*31/100dz 40 0 : [_e—3z/100]::=e—1.2; (b) Flaws occur randomly so we are observing a. Poisson process. 4.3A8 (a) W has a gamma distribution with a : 7, 6 f 1/16. 6 8ke‘8 (b) Using Table III in the Appendix, P(W _<_ 0.5) : 1 - k=U ' : 1 — 0.313 = 0.687, because here Am : (16)(0.5) :: 8. 4.3—14 Note that A = 5/10 = 1/2 is the mean number of arrivals per minute. Thus 9 z 2 and th p.d.f. of the waiting time before the eighth toll is f”) : ruins “He—“2 1 ' rt?) 216/3 ” 16/2—4 6"”, 0<z<oo, the p.d.f. of a chi-square distribution with r = 16 degrees of freedom. Using Table IV, P(X > 26.30) = 0.05. 4.4-4 (3.) 1.282; (b)—1.645; (c) —1.66; (d) >71.82. 4.4—16 (a) P(X > 22.07) : P(Z > 1.75) : 0.0401; (b) P(X < 2085?) = P(Z < —1.2825) = 0.10. Thus the distribution of Y is 005, 0.10) and from Table II in the Appendix, P(Y g 2) : 0.8159. 4.4718 (3.) P(2546.56 g X 3 3723.44) : P(—1.282 3 Z g 1.282) : 0.80; (b) P(2379.94 g X 5 3890.06) = P(—1.645 3 Z 5 1.645) = 0.90; (c) P(2067.37 g X 5 4202.63) = P(—2.326 3 Z 5 2.320) = 0.98; (d) P(X 3 2546.56) 2 P(Z g —1.282) : 0.10; Y is b(20,0.10) so P(Y g 3) : 0.8670. ...
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stat400sln6 - 4.2—2 p 03 U2 2 1/3 ﬂx —1.o"-o.s...

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