GRE 8768 Questions: 58 and 60

Forum for the GRE subject test in mathematics.
moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

GRE 8768 Questions: 58 and 60

Postby moo5003 » Thu Nov 06, 2008 6:31 pm

I just finished taking the GRE 8767 under a time limit and did rather well but would like to go over some of the problems I got wrong.

Question 58

If f(z) is an analytic function that maps the entire finite complex plane to the real line, then the imaginary axis must be mapped onto?

A) Entire Real Axis
B) A point
C) A ray
D) An open interval
E) The empty set

Answer is B, I guessed A

So my incorrect logic was considering e^x which does not satisfy all the conditions. Whats an easy way to see that it must be mapped to a point? Is the function a+bi -> a , an analytic function?

Question 60

A fair die is rolled 360 times. The probability that 6 comes up on 70 or more rolls is?

A) Greater then .5
B) Between .16 and .5
C) Between .02 and .16
D) Between .01 and .02
E) Lower then .01

Answer C, Guessed B

Firstly I ignored the 0 and was trying to find the probability of 7+ six's on 36 rolls. The best way I found out how to find the probability was to:

(36C7 + 36C8 + 36C9 + 36C10 + 36C11 + 36C12 + 36C13...) / 6^36

But I didnt want to spend my time trying to figure this out by brute force. What is a better way to find this? I just realized I could find the probability of rolling 6 or lower six's, though that still seems like alot of calculation.

After reading a bit about it apparently there is an approximation method though it looks a little time consuming.

sachem
Posts: 9
Joined: Wed Oct 29, 2008 8:01 pm

Postby sachem » Thu Nov 06, 2008 7:06 pm

Question 58:

f(z)=f(x+iy)=u(x,y)+i*v(x,y) where u and v are real valued. so, v(x,y)=0 for all x and y, since f(z) maps the entire complex plane to the real line.

Now use the fact that f(z) is analytic - this means that the cauchy equalities must hold : du/dx=dv/dy, and du/dy=-dv/dy

so du/dx = 0, and then u=h(y)+c for some function h of y.
and du/dy=0 so then u=g(x)+c for some function g of x.

Then u(x,y)=c for some constant c.

So, f(z)=c...everything goes to a single point. I am pretty sure this is correct but please correct me if I'm wrong. Note that given the real part or the imaginary part ( u or v ) of an *analytic* function, you can figure out the other part ( up to a constant ).

60)
The expected value of the outcome {6} is 1/6*360 = 60. (for a binomial distribution)
The standard deviation is sigma^2= npq=360*1/6*5/6=50. So the standard deviation is root(50) which is 5root2...so we can approximate this with a N(60, 5root2) distribution and instead find P(z>10/5root2) or P(z>root2)=1-P(z<root2)..

I think at this point you would have simply had to have memorized the fact that for the standard normal distribution, p(z<one standard deviation away) = .84 so p(z>one standard devation away) is .16. Funny enough, that is the number that they gave.

If you're curious, the numbers are P(z less than mean+0 standard deviations) is .5, 1 standard deviation is .84, 2 standard deviations is .98, and 3 is .999

Cheers, hope that helped.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Thu Nov 06, 2008 10:43 pm

How did you get npq for the variance formula? I cant seem to follow that step :/. I guess i'm used to doing this for continuous density functions.

sachem
Posts: 9
Joined: Wed Oct 29, 2008 8:01 pm

Postby sachem » Thu Nov 06, 2008 10:56 pm

It's a commonly used fact that for a binomial distribution, ie something where you have n trials each with either "success" (ie, a 6) or "failure" (ie, any other number), that if probability of success is p, (in this case 1/6), then:

expected value = n*p
variance=n*p*q where q=(1-p)=probability of failure




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