SOLUTION: The CIRA 2014 fact book (https://cira.ca/factbook/2014/index.html) states a typical Canadian watched an average of 291 videos per month.Assume that Doug Mackenzie is a typical Cana

Question 1020636: The CIRA 2014 fact book (https://cira.ca/factbook/2014/index.html) states a typical Canadian watched an average of 291 videos per month.Assume that Doug Mackenzie is a typical Canadian who watches an average of 291 videos in each thirty day period.
A.Calculate the probability that, in the next thirty days, Doug watches exactly 270 videos.
B.Calculate the probability that, in the next thirty days, Doug watches less than 270 videos.
C.Calculate the probability that, in the next thirty days, Doug watches exactly 280 or 281 or 282 videos.
D.Calculate the probability that, in the next fifteen days, Doug watches exactly 140 videos.
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
The CIRA 2014 fact book (https://cira.ca/factbook/2014/index.html) states a typical Canadian watched an average of 291 videos per month.Assume that Doug Mackenzie is a typical Canadian who watches an average of 291 videos in each thirty day period.
A.Calculate the probability that, in the next thirty days, Doug watches exactly 270 videos.
B.Calculate the probability that, in the next thirty days, Doug watches less than 270 videos.
C.Calculate the probability that, in the next thirty days, Doug watches exactly 280 or 281 or 282 videos.
D.Calculate the probability that, in the next fifteen days, Doug watches exactly 140 videos.

Solution:
The Poisson distribution applies to this problem, where λ=290, where
the pmf(probability mass function for a discrete distribution) is
P(k;λ)=((λ^k)*e^(-λ))/k!

and cdf (cumulative distribution function) is
F(k;λ)=ΣP(i;λ) for i=0 to floor(k).
Certain calculators are available to calculate the cdf without having to do the laborious sum.

The results apply to any 30-day period, independent of what happened in the past.

(a) k=270
P(k;λ)
=P(270,291)
=((λ^k)*e^(-λ))/k!
=((291^270)*e^(-291))/270!
=0.01116

(b) k<270
F(269,291)
=ΣP(i;λ) for i=0 to floor(k)
=ΣP(i;291) for i=0 to 269
=0.10267

(c) 280<=k<=282
P(280;291)+P(281;291)+P(282;291)
=0.01931+0.02000+0.02063
=0.05994

(d) The duration is halved, so the value of λ as well.
Use λ=291/2=145.5, k=140.
P(140;145.5)
=((λ^k)*e^(-λ))/k!
=((145.5^140)*e^(-145.5;))/140!
=0.03033

Note that even though 140/145.5=280/291, the answers in parts (d) and (a) are not the same, because with a smaller value of λ, the peak moves to the left and becomes taller, hence a higher probability.

To visualize this phenomenon, compare the pmf for various values of λ in, for example,
https://en.wikipedia.org/wiki/Poisson_distribution
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