A fisherman goes out fishing every day. If the day he goes out is a rainy day, t

Question

A fisherman goes out fishing every day. If the day he goes out is a rainy day, the total weight (in pounds) of fish that he catches follows an exponential distribution with mean 3. If the day he goes out is not a rainy day, the total weight of fish he catches is |Y|, where the random variable Y is Normally distributed with mean mu and variance sigma^2. For each day, the probability of raining is 0.4. (a) Suppose the probability that Y is at most 4 pounds is 0.1539, and the probability that Y is greater than 5 pounds is 0.67. Find the values of the mean mu and the variance sigma^2, and give your answers accurate to two decimal points. (b) Using your results in part (a), find the cumulative distribution function of the total weight of fish this fisherman catches on a random day. (You may give your final answer as a mathematical expression that involves the cumulative distribution function of a standard Normal distribution. Use the notation Psi(z) which represents the cumulative distribution function for a standard Normal distribution, i.e., Psi(z) = P(Z lessthanorequalto z) for Z ~ N(0, 1).) (c) Using your results in part (a) and part (b), find the probability density function of the total weight of fish this fisherman catches on a random day. (You may give your final answer as a mathematical expression that involves the probability density function of a standard Normal distribution. Use the notation Psi(z) which represents the density function of a standard Normal distribution, i.e., phi(z) =1/squareroot 2 pi e^-z^2/2.)

Explanation / Answer

1. Here P ( rain) = 0.4 and P( does n't rain) = 1 - 0.4 =0.6

Pr(x) = Probabitiy of catching x fish on rainy day = E(3)

Pnr(x) = Probability of catching x fish on rainy day = N ( µ, 2)

so by the question, he goes out on non rainy day so P(x<=4; µ, 2) =0.1539

and P (x>5; µ, 2) = 0.67

value of Z for the given probabilities 0.1539 and 0.67 is -1.2 and 0.44, respectively.

(4 - µ)/ = -1.2 and (5 - µ)/ = 0.44

(4 - µ)/((5 - µ) = -1.2/0.44 => µ = 4.73 and 2 = 0.37

(b) the cumulative probabiity distribution function

FX(x) = P(rain) * P ( number of fish catches in rain) + P(don't rain) * P ( number of fish catches in rain)

= 0.4(1 - e-x) + 0.6 (x) [ where (x) = P(X <= x; 4.73; 0.37) and = 1/3

(c) PDF pf the given CDF can be calculated by differentiating it

fX(x) = 0.4*(1/ 3) * e-x/3 + 0.6(x) [ Where (x) = P(x; 4.73; 0.37)]

fX(x) = 0.133e-x/3 + 0.6(x)

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