CHAPTER 88 THE BINOMIAL AND POISSON DISTRIBUTIONS EXERCISE Pag 97 1. Concrt blocks ar tstd and it is found that on avrag 7% fail to mt th rquird spcification. For a batch of nin blocks dtrmin th probabilitis that (a) thr blocks and (b) fwr than four blocks will fail to mt th spcification. Lt probability of failur to mt spcification p = 0.07 and probability of succss q = 0.9 (9)(8) (9)(8)(7) q+ p = q+ 9 qp+ qp + qp+... hnc By th binomial xpansion ( ) 9 9 8 7 2 6 (9)(8) (9)(8)(7) 0.9 + 0.07 = (0.9) + 9(0.9) (0.07) + (0.9) (0.07) + (0.9) (0.07) +... ( ) 9 9 8 7 2 6 = 0.5204 + 0.525 + 0.1061 + 0.0186 + which corrsponds to 0 1 2 failing to mt th spcification. (a) Probability that thr blocks fail to mt spcification = 0.0186 (b) Probability that fwr than four blocks fail = 0.5204 + 0.525 + 0.1061 + 0.0186 = 0.9976 2. If th failur rat of th blocks in Problm 1 riss to 15% find th probabilitis that (a) no blocks and (b) mor than two blocks will fail to mt th spcification in a batch of nin blocks. Lt probability of failur to mt spcification p = 0.15 and probability of succss q = 0.85 (9)(8) (9)(8)(7) q+ p = q+ 9 qp+ qp + qp+... hnc By th binomial xpansion ( ) 9 9 8 7 2 6 (9)(8) (9)(8)(7) 0.85 + 0.15 = (0.85) + 9(0.85) (0.15) + (0.85) (0.15) + (0.85) (0.15) +... ( ) 9 9 8 7 2 6 = 0.216 + 0.679 + 0.2597 + 0.1069 + which corrsponds to 0 1 2 failing to mt th spcification. (a) Probability that no blocks fail to mt spcification = 0.216 (b) Probability that mor than two blocks fail = 1 (0.216 + 0.679 + 0.2597) = 0.1408 170
. Th avrag numbr of mploys absnt from a firm ach day is 4%. An offic within th firm has svn mploys. Dtrmin th probabilitis that (a) no mploy and (b) thr mploys will b absnt on a particular day. Lt p = 4% = 0.04 thn q = 0.96 (i.. 96% prsnt) (7)(6) + = 0.96 + 0.04 = (0.96) + 7(0.96) (0.04) + (0.96) (0.04) ( q p) ( ) 7 7 7 6 5 2 which corrsponds to 0 1 2. mploys bing absnt. (7)(6)(5) (0.96) 4 (0.07)... + + (a) Th probability that no mploy will b absnt on a particular day = ( ) 7 0.96 = 0.7514 (b) Th probability that thr mploys will b absnt on a particular day (7)(6)(5) 4 = ( 0.96 ) ( 0.04 ) = 0.0019 4. A manufacturr stimats that % of his output of a small itm is dfctiv. Find th probabilitis that in a sampl of tn itms (a) fwr than two and (b) mor than two itms will b dfctiv. Lt p = % = 0.0 thn q = 0.97 (i.. 97% non-dfctiv) (10)(9) + = 0.97 + 0.0 = (0.97) + 10(0.97) (0.0) + (0.97) (0.0) ( q p) ( ) 10 10 10 9 8 2 (10)(9)(8) (0.97) 7 (0.0)... + + = 0.774 + 0.2281 + 0.017 + 0.0026 which corrsponds to 0 1 2. itms bing dfctiv. (a) Th probability that fwr than two itms will b dfctiv = 0.774 + 0.2281 = 0.9655 (b) Th probability that mor than two itms will b dfctiv = 1 (0.774 + 0.2281 + 0.017) = 0.0028 171
5. Fiv coins ar tossd simultanously. Dtrmin th probabilitis of having 0 1 2 4 and 5 hads upwards and draw a histogram dpicting th rsults. Lt probability of a had p H = 0.5 and th probability of a tail p T = 0.5 5 5 4 (5)(4) 2 (5)(4)() 0.5 + 0.5 = (0.5) + 5(0.5) (0.5) + (0.5) (0.5) + (0.5) (0.5) (5)(4)()(2) + 0.5 4! 0.5 + 0.5 ( ) = 0.01 + 0.156 + 0.125 + 0.125 + 0.156 + 0.01 which corrsponds to 0 1 2 4 and 5 hads landing upwards A histogram dpicting this data is shown blow ( )( ) ( ) 4 5 6. If th probability of rain falling during a particular priod is 2/5 find th probabilitis of having 0 1 2 4 5 6 and 7 wt days in a wk. Show ths rsults on a histogram. Lt probability of rain falling p = 2/5 = 0.40 and th probability of rain not falling q = 0.6 7 7 6 (7)(6) 5 2 (7)(6)(5) 4 0.6 + 0.4 = (0.6) + 7(0.6) (0.4) + (0.6) (0.4) + (0.6) (0.4) (7)(6)(5)(4) (7)(6)(5)(4)() (7)(6)(5)(4)()(2) + + + + 4! 5! 6! ( ) ( 0.6) ( 0.4) ( 0.6) ( 0.4) ( 0.6)( 0.4) ( 0.4) 4 2 5 6 7 = 0.0280 + 0.106 + 0.261 + 0.290 + 0.195 + 0.0774 + 0.0172 + 0.0.0016 which corrsponds to 0 1 2 4 5 6 and 7 wt days. 172
A histogram dpicting this data is shown blow 7. An automatic machin producs on avrag 10% of its componnts outsid of th tolranc rquird. In a sampl of tn componnts from this machin dtrmin th probability of having thr componnts outsid th tolranc rquird by assuming a binomial distribution. Lt th probability of a componnt bing outsid th tolranc p = 10% = 0.1 and th probability of a componnt bing within tolranc q = 0.9 10 10 10 9 (10)(9) 8 2 (10)(9)(8) 7 + = 0.9 + 0.1 = (0.9) + 10(0.9) (0.1) + (0.9) (0.1) + (0.9) (0.1) +... ( q p) ( ) which corrsponds to th probability of 0 1 2 componnts bing outsid th tolranc Th probability of having thr componnts outsid of th rquird tolranc 17
(10)(9)(8) 7 = ( 0.9 ) ( 0.1 ) = 0.0574 174
EXERCISE 4 Pag 99 1. In Problm 7 of Exrcis pag 97 dtrmin th probability of having thr componnts outsid of th rquird tolranc using th Poisson distribution. Avrag occurrnc of a componnt bing outsid of tolranc λ = np = (10)(0.1) = 1 Th probability of 0 1 2 componnts outsid of tolranc is givn by th trms of: λ λ λ 1 + λ + + +... = 1 1 1 1 1 1 + + + +... Th probability of thr componnts bing outsid of tolranc = 1 1 = 0.061 2. Th probability that an mploy will go to hospital in a crtain priod of tim is 0.0015. Us a Poisson distribution to dtrmin th probability of mor than two mploys going to hospital during this priod of tim if thr ar 2000 mploys on th payroll. Avrag occurrnc of th vnt λ = np = (2000)(0.0015) = Th probability of 0 1 2 mploys going to hospital is givn by th trms of: λ λ λ 1 + λ + + +... = + + + +... = 0.0498 + 0.1494 + 0.2240 + Th probability of mor than two mploys going to hospital = 1 (0.0498 + 0.1494 + 0.2240) = 1 0.422 = 0.5768. Whn packaging a product a manufacturr finds that on packt in 20 is undrwight. Dtrmin th probabilitis that in a box of 72 packts (a) two and (b) fwr than four will b undrwight. Probability of a packt bing undrwight p = 1 20 = 0.05 and n = 72 175
Hnc th avrag occurrnc of vnt λ = np = (72)(0.05) =.6 Th probability of 0 1 2 packts bing undrwight is givn by th trms of: λ λ λ 1 + λ + + +... =.6.6.6.6.6.6 +.6 + + +... = 0.027 + 0.0984 + 0.1771 + 0.2125 + (a) Th probability that two will b undrwight = 0.1771 (b) Th probability that fwr than four will b undrwight = sum of probabilitis of 0 1 2 and = 0.027 + 0.0984 + 0.1771 + 0.2125 = 0.515 4. A manufacturr stimats that 0.25% of his output of a componnt is dfctiv. Th componnts ar marktd in packts of 200. Dtrmin th probability of a packt containing fwr than thr dfctiv componnts. Avrag occurrnc of dfctiv componnts λ = np = (200)(0.0025) = 0.5 Th probability of 0 1 2 dfctiv componnts is givn by th trms of: λ λ λ 1 + λ + + +... = 0.5 0.5 0.5 0.5 0.5 0.5 + 0.5 + + +... = 0.6065 + 0.0 + 0.0758 + Th probability of a packt containing fwr than thr dfctiv componnts. = 0.6065 + 0.0 + 0.0758 = 0.9856 5. Th dmand for a particular tool from a stor is on avrag fiv tims a day and th dmand follows a Poisson distribution. How many of ths tools should b kpt in th stors so that th probability of thr bing on availabl whn rquird is gratr than 10%? Avrag occurrnc of dmand λ = 5 Th probability of 0 1 2 tools bing dmandd is givn by th trms: 176
λ λ λ 2 λ λ λ λ i.. 5 5 5 2 5 5 5 5 i.. 0.0067 0.07 0.0842 0.1404 0.1755 0.1755 0.1462 0.1044 0.065 Hnc th probability of wanting a tool ight tims a day is 0.065 i.. 6.5% which is lss than 10%. Thus svn tools should b kpt in th stor so that th probability of thr bing on availabl whn rquird is gratr than 10% 6. Failur of a group of particular machin tools follows a Poisson distribution with a man valu of 0.7. Dtrmin th probabilitis of 0 1 2 4 and 5 failurs in a wk and prsnt ths rsults on a histogram. Man valu λ = 0.7 Th probability of 0 1 2 4 and 5 failurs in a wk is givn by th trms: λ λ λ 2 λ λ λ λ λ 4 λ 4! and λ 5 λ 5! i.. 0.7 0.7 0.7 2 0.7 0.7 0.7 0.7 4 0.7 0.7 4! and 5 0.7 0.7 5! i.. 0.4966 0.476 0.1217 0.0284 0.0050 and 0.0007 A histogram dpicting ths rsults is shown blow: 177