Discrete Random Variables University of South Caroina离散随机变量南卡罗来纳州大学

1、020406080100120blk spots scratchessplaysinksflow lineotherfrequency01020304050607080topmidbottomleftrightfrequencyouncesnumber of active pumps16.0 16.2 16.4 let random variable y = number of aircraft in one day that arrive late. p(y=y)p(y=y)let y = number of components within specification in a rand

2、omly chosen assemblyy 0 1 2 3p(y=y) 0.00005 0.00635 0.14715 0.846451)(ypy 0 1 2 3p(y=y) 0.00005 0.00635 0.14715 0.84645y 0 1 2 3p(y=y) 0.00005 0.00635 0.14715 0.84645distribution function:y 0 1 2 3p(yy) 0.00005 0.00640 0.15355 1.0000cumulative distribution function:distribution functioncumulative di

3、stribution functionp(y = y)p(y y)y 0 1 2 3p(y=y) 0.00005 0.00635 0.14715 0.84645yyypye)()()()var(22yeyy 0 1 2 3p(y=y) 0.00005 0.00635 0.14715 0.84645yypyy)()()var(22)()var(22yey222)()var(yeyyyypyypyy222)()()var(2project a b cprob of accept 0.30 0.80 0.10assuming the projects are independent of one a

4、nother, what is the probability that the firm will have all three projects accepted?project a b cprob of accept 0.30 0.80 0.10 y 0 1 2 3 p(y) 0.126 0.572 0.278 0.024what is the expected number of projects accepted? y 0 1 2 3 p(y) 0.126 0.572 0.278 0.024what is the variance for number of projects acc

5、epted?)!( !rnrnrn“n choose r”ynyppynyp)1 ()(for y = 0, 1, 2, ,nnpye)()1 (2pnp)1 (pnp125025125025outcomes possible all of #occurcan a waysof #)(ap410022298410012398)21(ypnnynrnyryyp)(for y = 0, 1, 2,nnrnye)(10100109505910099505101009951510100095105abcd!)()(yetyptyfor y = 0, 1, 2,where is the average number of occurrences per base unit.and t is the number of base units inspected.further:2)(tye! 16 . 2! 06 . 2) 10(6 . 216 . 20eeyplet y = number of breaks in 1000 mete