For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains below a critical value y0 (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). An article proposes a geometric distribution with p = 0.369 for this random variable. (Round your answers to three decimal places.)

(a) What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals?

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

Question

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains below a critical value y0 (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). An article proposes a geometric distribution with p = 0.369 for this random variable. (Round your answers to three decimal places.)

(a) What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals?

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

anonymous · Answer

That's wrong =/  I have the solutions but they don't have the steps. I'm just trying to understand how they got there =/

anonymous · Answer

a)
exactly 3
 P( X = 3) =  (1- .369)^2 * .369 = .14692
at most 3
P ( X <= 3 ) = P(X = 1) + P( X = 2) + P( X = 3) 
= .369 + (1-.369)(.369) + (1- .369)^2 * .369 =

anonymous · Answer

They should be 0.093 and 0.841 :/

anonymous · Answer

you sure you copied the problem correctly and the solutions

anonymous · Answer

Yes.

anonymous · Answer

you want the drought to last 3 intervals. so you need to top off the 3 drought intervals with a surplus interval. 
label a drought interval with F. 
label surplus interval by S
P( FFFS) = ( 1- .369) ^3 (.369)

anonymous · Answer

Ahh okay that makes more sense

anonymous · Answer

the distribution we are using is 
P( X = k) = (1-p)^k*p ,  k = 0,1,2,3, ...

a) exactly 3 
P( X = 3) = (1- .369)^3 * .369

b) at most 3 
P ( X <= 3 )= P(X = 0) + P( X = 1) + P( X = 2)  + P(X=3)
=(1-.369)^0*(.369) + (1-.369)^1*(.369) + (1-.369)^2*(.369) + (1-.369)^3*(.369) 
= .841467

anonymous · Answer

Okay, I understand that so far

anonymous · Answer

you dont need to memorize the distribution formula
we know the success occurs on the fourth trial, since we want 3 consecutive drought intervals. ie P( FFFS)

anonymous · Answer

Okay so what about b?

anonymous · Answer

mean = ( 1- p) / p = (1-.369) / .369 = 1.71 
st. dev = sqrt[ ( 1- p) / p^2] = 2.1527
P ( Y >= 1.711 +2.1527) = P( Y >= 3.8627 )  = P( Y > 4)

anonymous · Answer

P(Y > 4) = 1 - P( Y <= 4) 
or we can save time using the formula:
P( Y >= k ) = (1 - p)^k 
P( Y > 4) = P( Y >= 5) = ( 1- .369)^5 = .1000

anonymous · Answer

The answer they gave is 0.159