WARM UPWARM UP

Eighty sixEighty six percent of seniors got percent of seniors got drunkdrunk on New on New

Year’s Eve night (hope you had fun at Steve Year’s Eve night (hope you had fun at Steve

Bart’s House!). There are seven seniors Bart’s House!). There are seven seniors

walking down the hallway. walking down the hallway.

1) Find the probability that 1) Find the probability that exactlyexactly four of them four of them

were were WASTEDWASTED. .

2) Find the probability that 2) Find the probability that at leastat least five of them five of them

were were WASTEDWASTED..

x n xn xC p q

Binomial Distribution Binomial Distribution formulaformula

p(p(successsuccess) = ) = pp

p(p(failurefailure) = ) = qq

nn = = totaltotal number of repeated trials number of repeated trials

xx = = selectedselected number of successes number of successes

x n xn xC p q

Section 6.6Section 6.6

Finding the MEAN, Finding the MEAN, VARIANCE, and VARIANCE, and

STANDARD DEVIATION of STANDARD DEVIATION of a Binomial Distributiona Binomial Distribution

= mean

n = total number

p = probability of success

MeanMean of the Binomial of the Binomial DistributionDistribution

x np

x

VarianceVariance and and Standard Standard DeviationDeviation

The variance of a binomial distribution is:The variance of a binomial distribution is:

The standard deviation of a binomial The standard deviation of a binomial distribution is:distribution is:

variance npq

. . variances d npq

DICE GAMESDICE GAMES1) A die is rolled 1) A die is rolled 600600 times. Find the times. Find the

meanmean, , variancevariance, and , and standard deviationstandard deviation

of the number of of the number of 44’s that are rolled.’s that are rolled.

x np

n = total number = 600p = p(success) = p(4) = 1/6

mean = 600(1/6) = 100

variance npqq = p(failure) = p(not 4) = 5/6

variance = 600(1/6)(5/6) ≈ 83.333

. . variances d npq 83.3333 9.1287

VACATION!!!!VACATION!!!!2)2) Liberty Travel finds that Liberty Travel finds that 7%7% of all vacation of all vacation

packages are canceled each year. If they packages are canceled each year. If they

estimate estimate 500500 packages a year, packages a year, how manyhow many can can

they expect will be they expect will be canceledcanceled (aka find the (aka find the

mean)mean)? ?

3)3) Find the Find the variancevariance and and standard deviaton.standard deviaton.

n = total number = 500p = p(success) = p(cancel) = 7%q = p(failure) = p(not cancel) = 93%mean = 500(7%) = 35variance = 500(7%)(93%) = 32.55

32.55 5.705