Question

Statistics - Please help! Thanks.

5. Let X be the random variable that describes the measurements of the diameter of Venus. We know that X is normally distributed with mean u = 7848 miles and standard deviation o = 310 miles. What is (a) P(x < 7000) (b) P(8000< x < 8100) (c) Verify your answers using R.

Answer 1

- 310 Anssia Since & N/7A81 X = N(7848, (310) 2) =) X-7848 N10,1) 2= 310 Hence P(X L1000) = P x-7848 < 7000-78481 310 310 ) = P(Z L-848) = P(Z <-2.735) = $(-2.720 = 1-0 (2.735) 2 0-997 =1-0.994 [ Using standaard - normal takh 0.003 and since 04-2)-T-(PZ) Ans-566) P (sooo <x<8100) - Plovo-7848 -1848 8100-7848 ) 319 310 2 ~ N(0,1) - as in part (a) = P(0.49 <2< 0813) = P(Z < 0.813) - P(Z <0.49) = $(0.813) - $ 0.49) - $ (6) is celf of standard = 0.792 -0.688 ( Using Standard normal variate Z.. (normal table - 310 310 Ans-s12 Verification for part la) Rcode: norm(7000, 7848,310) output! 0.0031 . Hence verified. R Code : phorm (8100, 748 phorm ($100, 1848, 310) - - phorm (8000, 7848, 310). output: 0.1038 That is approximately 0-104 Hence verified. Ang bra het X be the life time of roller burring criver that, x n Weibull with B-2, 8 - 7500 Hence of(X) = 2 2 2-1 = (2/7500)2 x30 - The probability that the bearing will last at hast jouw hours - PlX 78000)

= f I-PLX < 8000) in 8.oo =) P(X28000) = 1- ( f(x3 - Coo het Gotowe = + => 240.) se dae dt and as t zo =) t=0 . x=8000 = += fuery? 1.1378 we have 11-1378 1.1378 P ( XZ 8000) = 1-C e tat a I et] oo Hence =) Plxz poooh - to f-1,1378 +1] - T-04378 _ PC XZ 80000.86221 _ 2 2000 .86234 Ans Ans 6(6) Verification = P(X> 8000) = [P(x=8000) = 1 + 8 1.1378 – po 0.3205) Aus 6(6) Verification: R code: I t weibull (8000, shape = 2, scale = 7500) output: 0.3205305