I believe I am quite close to solving this, but I would just like to double check some of these answers.

Two species have different size toes. Lengths of toes of species X is normal distributed with mean 6 and variance 3. Toes of species Y is normal distributed with lengths $N(5,2)$. Species Y is more prevalent with .8 of occurrences compared to .2 for species X.

a) What is prob of finding toe belonging to one of these two types measuring more than 7 inches?

So, Let p=.8 for prob of finding species B. Therefore this is $p*P(Y>7)+(1-p)*P(X>7)$. Therefore we have $.8*(1-\phi(2/\sqrt(2))$+$.2*(1-\phi(1/\sqrt(3))$. Is this the correct formula and process? $\phi$ is the normal distribution z-score function

b) Suppose we find a toe of length greater than 7. What is prob belongs to X? belong to Y?

For both of them I said $P(X,Y>7)=1-P(X,Y\le7)$ where , means or in this case. So $P(X\le7)=\phi(\frac{7-6}{\sqrt(3)})$ and $P(Y\le7)=\phi(\frac{7-5}{\sqrt(2)}) $. Then solve for those and do 1 minus what we get.

c) Finally, these two species lived at different time. What is prob that next toe find is between 5 and 7 inches?

For this part I said find $P(5\le X,Y\le7)$ for both and then multiply by prob of find Y and prob of find X respectively. For prob of X I said this was $\phi(1/\sqrt(3)-[1-\phi(1/\sqrt(3)])=.438.$ Then for prob of Y I said this was $\phi(2/\sqrt(2)-\phi(0))=.4207$. Therefore multiply by respective probabilities and good to go. Correct? I am pretty sure about the normal random variable part in terms of the $\phi$ function, but just in case.


General comments. The distribution of toe lengths of the two species together is called a 'mixture' distribution. The mixture of two independent normal distributions is not normal (if means and variances differ), but may be difficult to distinguish from normal. If the means are sufficiently far apart relative to the variances, the mixture of two normals is bimodal, but this does not apply to your situation. The Wikipedia article on 'mixture distributions' shows an example of a bimodal distribution near the end. In your case the PDF is $f_{mix}(t) = p_1f_X(t) + p_2f_Y(t).$ This site has some related information on simulating mixture distributions at 70855; the R code there is correct and not wasteful of random digits, but much more complicated than necessary.

(a) Your notation and method in this part are not quite right. The usual notation for the standard normal CDF is $\Phi$ (not $\phi$, which is used for the density function). Also, $$P\{X > 7\} = P\{Z > (7-6)/\sqrt{3}\} = 1 - \Phi(1/\sqrt{3}) = 0.28185,$$ and similarly for $P\{Y > 6\}.$ With these adjustments, you should get the right answer, which I found by simulation to be about 0.12.

(b) Your explanation seems to have some words missing. Also, the notation $X,Y$ for the mixture random variable invites confusion; you should use a new random variable, say $W$ for the mixture distribution.

You can get the right answer using Bayes' Theorem. For the moment, letting $G = \{W > 7\}$, and letting $A$ and $B$ denote the species associated with random variables $X$ and $Y$, respectively, we seek $P(B|G) = P(B \cap G)/P(G),$ where the denominator was found in (a), and one term of it is the numerator. The answer is a little above 1/2.

(c) I don't understand why the question mentions two different periods of time. That would seem to be a clue that $X$ and $Y$ are independent random variables, but we have been making that assumption from the start.

If the question is simply asking for $P\{5 < W \le 7\}$, then a method similar to that of (a) would work, and I don't see that multiplication of probabilities is involved. From simulation I get about 0.42.

In case it might be of use to you, I'm putting the R code for my simulation below:

> m = 10^6;  p = rbinom(m, 1, .2)
> x = rnorm(m, 6, sqrt(3));  y = rnorm(m, 5, sqrt(2))
> w = p*x + (1-p)*y
> mean(w > 7);  mean(p[w > 7]==0);  mean((w > 5)&(w <= 7))
[1] 0.119877  # approx P(W > 7)
[1] 0.5285334 # approx P(B|Y > 7)
[1] 0.423371  # approx P(5 < W <= 7)

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