I am working on a Poisson distribution problem stated in the main question and got stuck and do not know how to proceed as I did not understand the next question on how to work it out
The following data collected from the Australian Bureau of Meteorology Website
(http://tinypic.com/r/2ibi9gi/8) gives the daily rainfall data for the year 2014 in Sydney in mm I have also calculated the mean and std deviation for each month
Can someone check my work ? for the part a (i) and (ii)
a) (i) What is the probability that on any given week in a year there would be no rainfall?(Poisson)
In 52 weeks, we 131 rainy days (mean summation for year 2014)
Therefore, in one week, it rain
γyear= 131/364, γwk= γyear/52
P(X=0)= (γ^X×e^(-γ))/γ!= (〖(γyear/52)〗^0 e^(-(γyear/52)))/0!=0.993
(ii)What is the probability that there will be 2 or more days of rainfall in a week? (There are 52 weeks in a year and a week is assumed to start from Monday). In a similar manner,
P(X=1)= (γ^X×e^(-γ))/γ!= (〖(γyear/52)〗^1 e^(-(γyear/52)))/1!=0.0069
With the mentioned formula, for
I did not understand how to proceed with the following part
(b) Assuming that the weekly total amount of rainfall from the data provided in part (a) has a normal distribution, compute the mean and standard deviation of weekly totals.
(i) What is the probability that in a given week there will be between 4 mm and 12 mm of rainfall?
(ii) What is the amount of rainfall if only 8% of the weeks have that amount of rainfall or higher?
thank you and much appreciated
as mentioned in the answer for question ii)
In 365 days, it rains 32.62206 mm (mean summation for year 2014) Therefore, in one day, it rains 32.62206/365
mean = 32.62206/365 Poisson distribution with mean 32.62206/365. P(no rain) = P(x=0) = e^-(32.62206/365) (32.62206/365)^0 /0! = e^(-32.62206/365) =P= 0.914 When using a Binomial with 7 trials x>=2 successes and a probability of 1-0.914 the answer obtained is P(X>=2) = 0.00349
from the following: http://stattrek.com/online-calculator/poisson.aspx