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leonid [27]
4 years ago
13

Find | P | a. 15 b. 22 c. 24 d. 8

Mathematics
1 answer:
ankoles [38]4 years ago
5 0

Answer:

c. 24

Step-by-step explanation:

15+9=24

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It is reasonable to model the number of winter storms in a season as with a Poisson random variable. Suppose that in a good year
erastova [34]

Answer:

E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4

Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24

Step-by-step explanation:

Previous concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

Other equations useful:

Let X and Y random variables

E(X) =E[E(X|Y)] (conditional expectation)

Var(X)=E[Var(X|Y)]+Var[E(X|Y)] (Total variance)

Solution to the problem

Let A the random variable that represent the number of winter storms next year

B a binary variable, B=1 if the next year is a good year and B=0 in the other case. Then we have this:

E(A|B=1) = 4  and E(A|B=0)=5

We can use the propoerties of conditional expectation like this:

E(A)= E[E(A|B)]= E(A|B=1)P(B=1) +E(A|B=0)P(B=0)

E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4

And we can use also the properties for conditional variance we have the following values:

Var(A|B=1)=4 Var(A|B=0)=5, by the propertis of the Poisson distribution

And then the conditional variance is givne by:

Var[E(A|B)]= E(A|B=1)^2 P(B=1) +E(A|B=0)^2 P(B=0)

And if we replace we got:

Var[E(A|B)]= 4^2 *0.6 +5^2 *0.4 =19.6

And we have also that the expected value for the conditional variance is given by:

E[Var(A|B)]= 4*0.6 +5*0.4 =4.4

And then finally the variance for the random variable A is given by:

Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24

5 0
4 years ago
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