The$Binomial$Model$$From$chapter$6.266.3$$à$normal$probability$model$applies$to$many$different$types$of$variables$(including$biological$measures$$à$normal$variables$are$numeric$and$continuous$à$when$variables$are$discrete$valued$numeric$variables,$such$as$counts$of$whole$numbers,$the$Binomial$Probability$Model$is$used$$$Background$Information$$àA$Bernoulli$trial$is$the$most$important$random$process$$ $ à$it$conists$of$INDEPENDENT$TRIALS$with$TWO$OUTCOMES$only$and$with$CONSTANT$PROBABILITIES$from$trial$to$trial$$$à$it$represents$the$mathematical$abastraction$of$coin$tossing.$The$Bernoulli$trial$leads$to$the$BINOMAL$probability$distribution.$This$is$illustrated$with$a$Galton$Board(Quineunx)$$$The$Galton$Board$(Quincunx)$à$Each$time$a$ball$hits$one$of$the$nails$it$can$bounce$either$right(probability$p)$or$left$(probability$16p)$à$for$symmetrically$placed$nails,$balls$have$an$equal$probability$of$bouncing$left$or$right$$à$if$the$rows$are$numbered$from$0$to$N,$the$path$of$each$falling$ball$is$a$Bernoulli$trial$consisting$of$N$steps$$à$THIS$PROCESS$GIVES$RISE$TO$BINOMIAL$DISTRIBUTION$$$$The$Binomial$Model$$à$this$model$applies$if$$• There$are$a$fixed$number$of$trials$$• Only$two$outcomes$are$possible(yes$or$no,$heads$or$tails)$• The$probability$of$success,$p,$is$the$same$for$each$trial$$• The$trial$are$independent$$Binomial$or$NOT?$Please$state$whether$the$following$statements$are$binomial$data$or$not$$$a)40$randomly$selected$college$students$were$asked$if$they$their$major$in$ordet$to$get$a$good$job.$$b)$35$randomly$selected$Americans$were$asked$what$countries$their$mothers$were$born$$c)$to$estimatet$he$probability$that$students$will$pass$an$exam,$the$professor$records$the$students$study$groups$success$on$the$exam$$Answers:$a)is$binomial$$$$$$$$$$$$$$$$$$$$b)not$binomial$because$there$are$more$than$two$possible$answrs$$$$$$$$$$$$$$$$$$$$c)$not$binomial$because$the$outcomes$are$not$independent$from$one$another$$$$Visualizing$the$Binomial$with$SOCR(remember$last$lecture?)$$$$Here$is$an$snap$shot$of$SCR$b(15,$.5)$$$here$is$b(100,$.3)$using$advanced$software$$$$$Binomial$Notation$$à$the$form$is$b(n,p,x)$$ à$n=trials$where$p=probability$of$a$success$and$x=specific$number$of$successes$observed$$$à$so$b(10,$.5,$6)$is$interpreted$as$10$trials$with$a$probability$of$success$5/10$(.5$or$50%)$and$observing$ectly$6$heads$$$àhow$would$you$interpret$100,$.5,$50?$$Answer:$100$trials$with$a$probability$of$success$of$50%$and$observing$exactly$50$heads$$$$Vocabulary$on$inequalities$$• Exactly$à$=$• Less$than$à$<$$• At$least$à!≥$• More$than$à!>$• At$most$à≤$$(note:$less$than,$and$at$least$and$more$than$and$at$most)$$$Computing$Binomial$Probability$$à$12%of$all$US$women$will$eventually$develop$breast$cancer$• If$30$women$are$randomly$selected,$what$is$the$probability$that$exactly$4$of$them$will$eventually$develp$breast$cancer$$ B(30,0.12,$4)$$$$$$$$à$Remember:$n=trials$where$p=probability$of$a$success$and$x=specific$number$of$successes$observed$$$Using$SOCR$to$solve$$• Functions$à$PDF/CDFà$Binomial$$• N=30,$p=.12,$left$cut$off$=4$right$cut$off$=4$$à$the$probability$that$$exacly$4$of$the$30$women$will$develop$breast$cancer$is$about$.2046$$Here$is$some$information$of$SOCR$$$Finding$Binomial$Probabilities$$Table$3$in$appendix$A$of$the$text$book$can$be$used$to$find$basic$binomial$Probabilities$for$n=2$to$n=15$and$a$small$set$of$possible$p$à$we$can$also$use$this$probability$formula($found$at$the$top$of$page$273)$for$greater$flexibility$in$problem$solving.$$! = ! =!!!! ! − !(!)!(1 − !)!!!$$Binomial$Probability$by$Formula$$à12%$of$all$US$women$will$eventually$develop$breast$cancer$$• If$30$women$are$randomly$selected,$what$is$the$probability$that$exactly$4$of$them$will$eventually$develop$breast$cancer$$$à$b(30,$0.12,$4)=?$$Use$the$formula$! = ! =!!!! !!!(!)!(1 − !)!!!$First$Plug$in$the$numbers$$! = ! =30!4! 30 − 4(.12)!(1 − .12)!"!!$Begin$by$simplifying$$$$The$!$means$Factorial,$the$product$of$all$positive$integers$from$n$down$to$1(meaning$4!$Would$be$4*3*2*1)$So:$ $$Keep$simplifying:$$$Until:$$$$$Remember$this$from$last$week?$à$if$1/3$of$the$population$is$Hispanic$and$20$are$randomly$selected$people$from$the$population$are$chosen,$then$the$probability$distribution$function$for$the$number$of$Hispanics$is:$$$This$is$the$same,$only$how$we$are$being$specific$about$the$value$of$x.$With$this$we$would$answer$the$question:what$is$the$chance$of$getting$exactly$6$Hispanics$in$20?$$$$Finding$a$Binomial$probability$$à$14%$of$all$clothing$brought$online$is$returned.$$• If$an$online$retailer$sells$35$items$of$clothing$what$is$the$probability$that:$Questions$(try$it$on$your$own$first$then$check$your$answers$)$A) at$least$4$will$be$returned?$B) $Fewer$than$(or$less$than)$7$will$be$retu$.5540rned?$C) More$than(or$greater$than)$6$will$be$returned?$D) At$most$4$will$be$returned?$$The exclamation point means Factorial, the So:UntilThis is the same, only now we are being$$Answers$A) at$least$4$will$be$returned?$a. n=35$p=.14$P=(! ≥$5)$≈$.5540$B) $Fewer$than$(or$less$than)$7$will$be$returned?$a. ! = 35, ! = .14, ! ! < 7 ≈ .7892 $C) More$than(or$greater$than)$6$will$be$returned?$a. ! = 35, ! = .14, ! ! > 6 ≈ .2107$D) At$most$4$will$be$returned?$$a. ! = 35, ! = .14, ! ! ≤ 4 ≈ .4459$$$Why$are$we$doing$this:Given$that$we$can$calculate$the$probability$based$upon$information$about$a$population$(all$clothing)$we$can$use$this$to$check$the$performance$of$say$an$individual$retailer$or$a$group$of$retailers$this$leads$us$in$to$the$next$concept$$$Center$and$spread$$• the$binomial$has$a$mean$and$a$standard$devation$like$all$other$distributions$$• the$mean$is$where$the$distribution$balances$• the$standard$deviation$tells$us$how$far$the$distributions$are$from$the$mean$$• if$a$binomial$is$symmetric$then$the$mean$will$be$at$the$center$of$the$distribution$$• if$a$binomial$is$right$skewed$the$mean$will$be$to$the$right$of$the$peak$$• if$a$binomial$is$left$skewed$then$then$mean$will$be$to$the$left$of$the$peak$$Formulas$for$Center$and$Spread$$• For$a$binomial$distribution$! = !"$is$called$the$mean$or$expected$value$$• A$binomial$distribution$with$!$trials$and$a$probability$of$a$success$!$has$a$standard$deviation$of:$$•