normal distributionstatistics
I'm given the following equation: $P(X<b)=0.05$ and $X \sim N(-1, 4)$. Now I have to find the value of b in this equation. First, I convert $P(X<b)$ to be using the standard normal distribution. This gives $P(Z<\frac{b – \mu} \sigma) = 0.05$ or $P(Z<\frac{b + 1} 2)=0.05$. By looking into a table with all values for $\phi(z)$, I found out that $\frac{b + 1} 2 = -3.29$ or $b = -7.58$. The solution that was given by the teacher is $-4.29$.
I don't know what I did wrong?
Using a standard normal table, which you can find a good example of one here: https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf,
for $P(x < 31.5) = 0.05$, that means the area to the left of the z-score is $0.05$. Because we know that the distribution is normal, we would go into the middle of the table and find the closest value to $0.05$ as possible. So if we observe $z_1 = -1.64$ and $z_2 = -1.65$, the given areas to the left are $0.05050$ and $0.4947$ respectively.
To get a value as $close$ to $0.05$ as possible, we can interpolate by taking the midpoint of the two values. So $z_m = {1\over2}{(z_1 + z_2)} = -1.645$.
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))
Best Answer
If you look at this table you will seee that $\Phi(-1.645)=0.05$
Thus the equation is $\frac{b+1}{2}=-1.645$
I used linear interpolation. The mean of $\Phi(-1.64)$ and $\Phi(-1.65)$. This is almost $0.05$
Therefore $b=-4.29$