I prefer the computer type answer than hand writing. If you do hand writing, ple
ID: 3206585 • Letter: I
Question
I prefer the computer type answer than hand writing.
If you do hand writing, please make it sure your hand writing is readable.
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Remember our Minneapolis weather chain S 0.6 0.4 IN 0.2 0.8 where S stands for "snow" and N stands for "no snow". (a) on the long run, what is the probability that a given day is snowy, but the following day is not? (b) Assume it snowed today. How many days should we wait on average until it snows again?Explanation / Answer
a) P(following day is not snowy .Given a day is snowy) = 0.4
b)
1*(0.2)+2*0.8*(0.2)+3*0.82*(0.2)+.......... = 1/0.2 = 5
Expected(no of daysto snow gain) = 1*(0.6)+2*(0.4)*(0.2)+3*(0.4)(0.8)(0.2)+4*(0.4)(0.8)(0.8)(0.2)+...
= 0.6+(0.4)*(2*(0.2)+3*0.8*(0.2)+4*0.82*(0.2)+.........)
= 0.6+(0.4)*[(1/0.8)(2*0.8*(0.2)+3*0.82*(0.2)+...........)]
= 0.6+1/(2)[-1*0.2 +1*(0.2)+2*0.8*(0.2)+3*0.82*(0.2)+...........]
= 0.6+1/2[-0.2+1/0.2] = 0.6+(1/2)(-0.2+5) = 0.6+2.4 = 3