# according to climatologists, the long term average for atlantic storms is 9.6 per season (june1 to nov.30)...Question continues in additional notes...with 6 becoming hurricanes and 2.3 becoming...

according to climatologists, the long term average for atlantic storms is 9.6 per season (june1 to nov.30)...Question continues in additional notes...

with 6 becoming hurricanes and 2.3 becoming intense hurricanes. find the probability of the following events.

a) ten or more atlantic storms

b) five or fewer hurricanes

c) three or more intense hurricanes

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There are 9.6 atlantic storms in the season (Jun1 - Nov30) on average.

Assuming storms can only occur once a day and last only one day the estimated probability of a storm on any day in the season is

`hat p` = 9.6/(30 + 31 + 31 + 30) = 9.6/122 = 0.079

where the number of events x = 9.6 and the number of days n = 122

The number of atlantic storms X in a month follows a binomial distribution

`Pr(X=x) = (n!)/(x!(n-x)!) p^x(1-p)^(n-x)`

The probability of at most x storms is

`Pr(X<=x) = Sigma_(r=0)^x(n!)/(r!(n-r)!)p^r(1-p)^(n-r)`

So the probability of at least 10 storms is 1 - probability of at most 9 storms

`1-Pr(X<= 9) = 1-(Pr(X=0) + Pr(X=1) ... + Pr(X=9)) = `

`1- 0.506 = 0.494`

Similarly, the estimated probability of a hurricane is

`hat p = 6/122`

So the probability of five or fewer hurricanes is

`Pr(X <=5) = Pr(X=0) + Pr(X=1) ... + Pr(X=5) = 0.442`

Similarly, the estimated probability of an intense hurricane is

`hat p = 2.3/122`

So the probability of three or more intense hurricanes = 1 - the probability of at most two intense hurricanes is

`1- Pr(X<=2) = 1- (Pr(X=0) + Pr(X=1) + Pr(X=2)) = `

`1 - 0.595 = 0.405`

**a) 0.494**

**b) 0.442**

**c) 0.405**