Poisson Distribution

Formular PPT Page 67

Practice: Cookies quality, average number of chocolate- chips per cookie is 6

1) Pick a random cookies probiliy is fewer than 5 chips

Succes: Finding a chip; X: the number of successes; there is no a priority limit for X(Different to the size of bionatial distribution); lamda = 6 chocolete-chips per cookies;

Solution: P(X<5) = P(X <= 4) = F(4)

R Coding:

ppois(4, lambda = 6)

2) Pick a random cookies probiliti exactly 5 chips

R Coding:

dpois(5, lambda = 6)

3) 80% of the cookies will have at most what number of chips?

Want to find "b" such that P(X<= b) = 0.80

R Coding:

qpois(0.8, lambda = 6)

That means 80% of the cookies will have 8 or fewer chips; or 80% of the cookies will have at most 8 chips

4) 20% of the cookies will have at least what number of chips?

Solution: P(X>=c) =P(X>c-1) = 1 - F(c-1) = 0.2, so the ans: F(c-1) = 0.8 means 80% quantile is 9,so C=9

qpois(0.8, lambda = 6)

20% of the cookies will have at least 9 of chips?

5) Acookie is considered high quality, if it has at least 5 chips, otherwise, it will be disorded.

In a betch of 100 cookies, how many expected to be discarded?

Analysis: Y = numbers of discorded cookies out of the 100 in the betch; Maximum for Y is 100; success-> cookie with fewer than 5 chips -> disorded.

Y is Binomial n=100, we need Pi to caculate. Pi = P(X<5) = P(X<=4) = F(4) = 0.285 (X is the number of chips on one cookie)

R coding

ppois(4,lambda = 6)

we got pi = 0.285, So the expected value of that is E[Y] = nPi = 100(0.285) = 28.5