The chances a customer will make a purchase is 60%. then the probability of 5th purchase in 12th customers is
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Recognizing Probability Distribution:
a)Poisson Distribution (average)
b)Binomial Distribution (winning 0r Losing)
c)Binomial Distribution (buying or not buying)
d)Normal Distribution (mean and sd)
Answering the various examples of Probability Distributions:
a)P(x > 7) = 1 - poissoncdf(5,7) = .1334 0r 13.34%
b)p(winning) = .55, n = 3, P(x = 1) = binompdf(3, .55,1)= .3341, .3341/(3C1) = .1114, winning on 3rd try
c)p(purchase) = .60, n = 12, P(x=5) = binompdf(12, .60, 5)= .1009/(12C5), 12th customer makes 5th purchase
d)P(22 < x < 28) = nomalcdf(22,28,25,3)
nomalcdf(22,28,25,3)x30days is the number of days
a)Poisson Distribution (average)
b)Binomial Distribution (winning 0r Losing)
c)Binomial Distribution (buying or not buying)
d)Normal Distribution (mean and sd)
Answering the various examples of Probability Distributions:
a)P(x > 7) = 1 - poissoncdf(5,7) = .1334 0r 13.34%
b)p(winning) = .55, n = 3, P(x = 1) = binompdf(3, .55,1)= .3341, .3341/(3C1) = .1114, winning on 3rd try
c)p(purchase) = .60, n = 12, P(x=5) = binompdf(12, .60, 5)= .1009/(12C5), 12th customer makes 5th purchase
d)P(22 < x < 28) = nomalcdf(22,28,25,3)
nomalcdf(22,28,25,3)x30days is the number of days
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