Question (a) The following data collected from the Australian Bureau of Meteorol

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(a) The following data collected from the Australian Bureau of Meteorology Website (http://www.bom.gov.au/climate/data/?ref=ftr) gives the daily rainfall data (includes all forms of precipitation such as rain, drizzle, hail and snow) for the year 2016 in Adelaide, South Australia. The zero values indicate no rainfall and the left-most column gives the date. Assuming that the weekly rainfall event (number of days in a week with rainfall) follows a Poisson distribution (There are 52 weeks in a year and a week is assumed to start from Monday. The first week starts from 4 January 2016 – you are expected to visit the website and get the daily values which are not given in the table below. Part of the 52nd week runs into 2017.

(i) What is the probability that on any given week in a year there would be no rainfall?

(ii) What is the probability that there will be 2 or more days of rainfall in a week?

(b) Assuming that the weekly total amount of rainfall (in mm) 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 8mm and 16mm of rainfall?

(ii) What is the amount of rainfall if only 12% of the weeks have that amount of rainfall or higher?

Station: Adelaide Airport Number: 23034 Opened: 1955 No: Open Lat: 34.95 S Lon: 138.52 EElevation: 2 m Key: Units = mm 12.3 Not quality controlled. = Part of accumulated total Show in table Move mouse over rainfall total to view the period of accumulation. 2016 Jan Feb Mar May Jun 0.2 0 0 9.2 10.8 2nd 3rd 4th 5th 6th 11.0 32.8 6.8 11.6 8th 7.8 2.4 0.2 11.8 0 5.8 18.4 10th 11th 12th 13th 14th 15th 16th 17th 18th 19th 20th 21st 22nd 23rd 24th 25th 26th 19.6 0.6 0.6 0.6 13.2 12.2 2.2 0 0.2 0.2 9.8 0 0.6 0.2 0.6 0.8 11.6 0.8 9.0 0 11.8 0.6 2.2 0.8 11.2 3.8 0 12.2 2.6 14.0 0.8 0 0 0.8 22.8 0.8 5.4 14.4 8.0 53.2 28th 29th 30th 31st Highest Daily Monthly Total 0.2 2.4 38.0 12.2 0.2 12.2 36.2 0.6 0.8 22.8 3.0 32.8 0.8 11.6 0 53.2 8.0 64.272.095.4 49.0 102.2 59.630.2 75.8 View all monthly data Plot year of daily data 14.0 11.8 19.6 41.6 10.8 38.0 13.2 14.8

Explanation / Answer

(A)  (i) What is the probability that on any given week in a year there would be no rainfall?

First we have to find that on any given week there would be no rainfall.

Number of such weeks = 5

so probability that there is no rain whole week = 5/52

(ii) What is the probability that there will be 2 or more days of rainfall in a week?

probability of NUmber of such weeks where there is 2 or more days of rainfall in a week = 1 - P(no rainfall in the week) - P(rainfall only on one day is a week)

P(no rainfal in a week) = 5/52

SO number of days when only rainfall on one day>

so Pr( rain occur 2 or more days) = 1 - (5 + 10)/ 52 = 37/52

(b) There are number of weeks = 52

Total rainfall = 649.0 mm

so Avergae rainfall per week = 649/52 = 12.48 mm

To calculate rainfall eachweek rainfall is calculated

Standard deviation of the average rainfall per week = 15.7 mm

(i) What is the probability that in a given week there will be between 8mm and 16mm of rainfall?

Pr (8 mm < rainfall per week < 16 mm)

=   (Z2 ) -  (Z1 )

where   is the cumulative normal distribution function.

Z2 = (16 - 12.5) / 15.7 = 0.22

Z1 = (8.0 - 12.5)/ 15.7 = -0.29

Pr (8 mm < rainfall per week < 16 mm) =   (Z2 ) -  (Z1 ) = (0.22) - (-0.29)

= 0.5871 - 0.3859 = 0.2012

(ii) LEts say that rainfall is X mm per year

so Pr( Rainfall >= X ; 12.5 ; 15.7) = 0.12

from Z - table checking the value of Z for p - vlaue= 0.12 is

Z = 1.175

( X -12.5)/ 15.7 = 1.175

X = 12.5 + 1.175 * 15.7 = 30.95 mm or say 31 mm

Week Rainfall 1 0 2 1.2 3 16.4 4 18.6 5 13.8 6 0 7 0.6 8 0.4 9 0 10 31.8 11 9 12 0.8 13 0 14 4.4 15 0.2 16 0.6 17 3.2 18 6.6 19 14.6 20 0.2 21 41 22 7.8 23 20.6 24 4.2 25 33.2 26 11.4 27 44.2 28 7.8 29 5.4 30 34.2 31 10.8 32 12 33 55.2 34 0.6 35 11 36 21 37 25.2 38 6.4 39 48.6 40 20.4 41 3 42 33.2 43 0.4 44 1 45 25.4 46 2.6 47 2 48 0 49 14.8 50 0.6 51 0.8 52 59.6

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