Question

craft breweries that make beer in small batches are experiencing a spectacular growth in bars and liquor stores across the nation. The craft beer industry now boasts of 4269 breweries, representing a 12% market share of the total beer market in the united states (Fortune, March 22, 2016). It has been estimated that 2 craft breweries open every day. Assume this number represents an average that remains the constant over time. a) What is the probability that exactly 4 craft breweries open in a day? b) What is the probability that there are at least 5 craft breweries open in a day? c) What is the probability that exactly 10 breweries open every week? d) What is the probability that at least 20 craft breweries open every week?

Answer 1

Solution:

We are given λ = 2 for a day.

We have to use Poisson distribution for finding the required probabilities.

P(X=x) = λ^x*exp(-λ)/x!

a) What is the probability that exactly 4 craft breweries open in a day?

Here, we have to find P(X=4)

P(X=4) = 2^4*exp(-2)/4!

P(X=4) = 16* 0.135335/24

P(X=4) = 0.090223

Required probability = 0.090223

b) What is the probability that there are at least 5 craft breweries open in a day?

Here, we have to find P(X≥5)

P(X≥5) = 1 – P(X≤4)

P(X≤4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

P(X=0) = 2^0*exp(-2)/0! = 0.135335

P(X=1) = 2^1*exp(-2)/1! = 0.270671

P(X=2) = 2^2*exp(-2)/2! = 0.270671

P(X=3) = 2^3*exp(-2)/3! = 0.180447

P(X=4) = 2^4*exp(-2)/4! = 0.090224

P(X≤4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

P(X≤4) = 0.947347

P(X≥5) = 1 – P(X≤4)

P(X≥5) = 1 – 0.947347

P(X≥5) = 0.052653

Required probability = 0.052653

c) What is the probability that exactly 10 breweries open every week?

Here, we have to find P(X=10)

We have λ = 2 for a day.

So, λ for week = 7*2 = 14

P(X=10) = 14^10*exp(-14)/10!

P(X=10) = 0.066281843

Required probability = 0.066281843

d) What is the probability that at least 20 craft breweries open every week?

Here, we have to find P(X≥20)

λ for week = 7*2 = 14

P(X≥20) = 1 – P(X≤19)

P(X≤19) = 0.923495

(by using Poisson table or excel)

P(X≥20) = 1 – P(X≤19)

P(X≥20) = 1 – 0.923495

P(X≥20) = 0.076505

Required probability = 0.076505