Proof, modular arithmetic, probs & stats

Hey, too many questions.

5. What is that

$\displaystyle \Phi^{-1}$

symbol about? How can I find these values using the z tables?

No, the reasoning is a bit faulty - I think you've mixed up where the "not" goes. You can tell it's faulty because it's true that if mn is even then m is even or n is even.

You wanted "mn even => m even or n even" has contrapositive "m odd and n odd => mn odd".
If you're asked to prove that mn is even implies m even or n is even, I think you need a special case of Bézout's Theorem,which states that if hcf(m,n) = k then am+bn = k has solutions. This adapts by:
If m is not even, then hcf(2, m) = 1 and hence am + 2b = 1 has solutions, and hence amn + 2bn = n has solutions. But mn is assumed to be even, and hence amn is, so the LHS is even, and hence the RHS is even, so n is even. Hence at least one of m and n is even.
Your answer on slide 2 is fine.

Mx(t) is not negative because the integrand is positive (exponentials are positive).

Your Venn diagram is fine. P(C U D) = P(we get one of b,e,d,f); you can look at this as P(we don't get a or c).

The inverse-phi is exactly the inverse function of the phi function; that is, if phi(0) = 0.5, then phi^-1(0.5) = 0. You use the phi tables "in reverse" - that is, instead of looking at the row, and following along the columns to reach the right entry, you look for the entry in the table, then find which row/column it lies in.

Hmmm. I thought it should be lambda over t - lambda?

So am I assuming that all events are contained in the sets A B C and D? P(C u D) = 0.5 ? Or am I making false assumptions that the sample space = 1? Maybe I have overcomplicated things

Ah, I misunderstood you; I thought you meant Mx(t) was negative, which in hindsight was a silly thing for me to think given that you wrote "the integral of Mx(t)" :P I suspect it's only defined for t>lambda, but I have forgotten the circumstances as the picture seems to have vanished!

Yep, the fact that it's a sample space means every possible outcome is in the sample space, so there are six possible outcomes and their total probability is 1. C U D together rules out four of the six outcomes (if the outcome were e, then the outcome is in C U D, etc). To check, I made P(C) = 3/10. We can see that all possible outcomes are contained somewhere in A, B, C and D, because every one of the six possible letters is used somewhere.

It's the inverse of the cumulative normal distribution.

You can either, for standard values, look it up in the small table that usually follows the standard normal distribution table,

or, look for the value you're interested in in the body of the normal distribution table, and see what z values gives that.

I edited my embarrassing proof out and must have taken that pic down, it is back now

Is this right?

$\displaystyle 3^{3^{3^{3}}}\equiv -72 \equiv 5\mod \ 11$

Hey, am still uncertain, oh, think it just clicked, so I was supposed to be looking for the value that gives 0.45 as the rhs of the distribution is full? I was looking at 0.475

OK.

It's more usual for tables to give the cumulative distribution from -infinity, rather than from 0, as the values will differ by 0.5 in the body of the table.

As long as you're happy that such things are taken care of if/as necessary in your calculations.

D Franklin and ghostwalker in one thread

I'll do the easy bit!

3. How is integral of Mx(t) not negative?

What restrictions do you have on t? Look at the form of $M_x(t)$ given to you after its been integrated. What do you notice?

t < lambda

Confused, I will show what am doing then maybe it will be something ridiculous as usual

$\displaystyle \lambda \int e^{(t- \lambda)x} \ dx = \frac{\lambda \cdot e^{(t- \lambda)x}}{t- \lambda}$

Where do I go from here to get that equation in the pic? Or is it just something I am supposed to take as given?

Evaluate it with the limits and then hopefully it'll make sense

had to look at evaluating integrals with infinity, so I get this



$\displaystyle = \frac{\lambda \cdot e^{-\infty} - \lambda}{t- \lambda}$

Is this right? If so I think I may just about get it