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A fi sherman 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), and the cumulative distribution function of the total weight of fi sh 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 $\Phi$(z) which represents the cumulative distribution function for a standard Normal distribution, i.e., $\Phi$(z) = P(Z $\le$ 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 $\phi$(z) which represents the density function of a standard Normal distribution, i.e., $\phi$(z) = $\frac{1}{\sqrt{2\pi}}e^{\frac{-z^{2}}{2}}$.)

So I did part a). I had $$ P(Y\le4)= P(\frac{Y-\mu}{\sigma} \le \frac{4-\mu}{\sigma}) = P(z \le \frac{4-\mu}{\sigma}) = 0.1539$$ $$ P(Y>5)= P(\frac{Y-\mu}{\sigma} > \frac{5-\mu}{\sigma}) = P(z > \frac{5-\mu}{\sigma}) = 0.67$$ So $$ \frac{4-\mu}{\sigma} = -1.02$$ $$ \frac{5-\mu}{\sigma} = -0.44 $$ I solved for $\mu$ and $\sigma$ and got $\sigma$ = 1.7241 and $\mu$=5.7586. I have no idea how to do part b) and c). Any help is appreciated. Thanks!

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Outline:

$1.$ First, your method for (a) is correct, and I tried verifying that your $\mu$ and $\sigma$ work. Probabilities from R statistical software are almost exactly correct, so your $\mu$ and $\sigma$ are about as close as you can possibly get using printed normal CDF tables.

pnorm(4, 5.7586, 1.7241)
## 0.1538618
pnorm(5, 5.7586, 1.7241)
## 0.3299694

$2.$ (a) The next logical step is to figure out the CDF for the catch on a day when it does not rain. Almost none of the probability of $\mathsf{Norm}(5.7586, 1.7241)$ lies below 0, so $|Y|$ is almost the same as $Y.$ The very small bit of the left tail of the distribution of $Y$ gets 'folded over' to become positive. (So little, that I'm wondering if you are just supposed to ignore the folding.)

pnorm(0, 5.7586, 1.7241)
## 0.0004187992

(b) From there, you need to take the appropriate 0.4:0.6 weighted average of the exponential and (almost) normal CDFs.

$3.$ Finally, you need to take the derivative of the 'mixed' CDF to find the 'mixed' PDF.

Addendum (per Comment). I like to check (and even anticipate) analytic results using simulation in R statistical software. Of course, a simulation doesn't 'prove' anything, but I think your CDF is OK.

In the simulation below, $W$ is $1$ for 'rain' and $0$ otherwise. $X$ is your exponential random variable (rate 1/3 to get mean 3), and $Y$ is the normal distribution with the mean and variance you found. In R pnorm (without mean and variance parameters) is standard normal CDF $\Phi.$

The empirical CDF (ECDF) of a sample of size $n$ jumps up by $1/n$ at each (sorted) observation. It is a good estimate of the population CDF, in the somewhat the same sense as a histogram of a sample estimates the population PDF (only better).

The dotted red line uses your CDF. (It is plotted over the ECDF, with a perfect match within the resolution of the graph) When you do part (c), you can check how well you PDF matches the histogram.

m=10^5;  w = rbinom(m, 1, .4);  x = rexp(m, 1/3)
mu = 5.7586; sg = 1.7241;  y = abs(rnorm(m, mu, sg))
catch = w*x + (1-w)*y
mean(x);  mean(y);  mean(catch); .4*mean(x)+.6*mean(y)
## 3.004829  # sim E(X) = 3
## 5.754262  # sim E(Y) = 5.7586 
## 4.663314  # sim E(Catch)
## 4.654489  

par(mfrow=c(1,2))
  hist(catch, prob=T, br=60, col="skyblue2")
  plot(ecdf(catch))
    curve(.4*pexp(x, 1/3)+.6*(pnorm((x-mu)/sg) - pnorm((-x-mu)/sg)), 0, 50,
       lwd=3, col="red", lty="dashed", add=T)
par(mfrow=c(1,1))

enter image description here

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  • $\begingroup$ So I got the CDF for rainy day as 1-$e^{\frac{-1}{3}x}$ and for the CDF of |Y|, I got P(-y$\le$Y$\le$y)=P(z$\le \frac{y-\mu}{\sigma}$)-P(z$\le \frac{-y-\mu}{\sigma}$)= $\Phi(\frac{y-\mu}{\sigma})$-$\Phi(\frac{-y-\mu}{\sigma})$. So then the CDF of the whole thing would just be 0.4(CDF of rainy day)+0.6(CDF of |Y|). Does this look right to you? @BruceET $\endgroup$ – gofish Mar 13 '17 at 7:22
  • $\begingroup$ Your "CDF" for $|Y|$ doesn't seem to reach 1 toward the right. Maybe add, not multiply? See Addendum. $\endgroup$ – BruceET Mar 13 '17 at 18:24
  • $\begingroup$ I actually have minus instead of multiplication. 0.6*(pnorm((x-mu)/sg) - pnorm((-x-mu)/sg)) @BruceET $\endgroup$ – gofish Mar 13 '17 at 20:10
  • $\begingroup$ Got my microscope out, and I see it now. Remember I mentioned how small the second term has to be. Changing the simulation to remove black line. $\endgroup$ – BruceET Mar 13 '17 at 20:52
  • $\begingroup$ Thanks for the help! Do you think you help me with another question? @BruceET $\endgroup$ – gofish Mar 14 '17 at 23:22

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