weilan8 wrote:#60 A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is
(A) greater than 0.50
(B) between 0.16 and 0.50
(C) between 0.02 and 0.16
(D) between 0.01 and 0.02
(E) less than 0.01
Answer:C
diogenes wrote:weilan8 wrote:#60 A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is
(A) greater than 0.50
(B) between 0.16 and 0.50
(C) between 0.02 and 0.16
(D) between 0.01 and 0.02
(E) less than 0.01
Answer:C
For this one it looks susceptible to transforming it into a standard normal random variable (we have large N here).
Set X to the "the number of times six is observed" and let N[*] be the Normal cdf.
So, we find that Pr[X>=70] =1-Pr[X<70] = 1- N[(70-mu)/sigma], with mu = (1/6)*360 = 60 and sigma =sqrt((360/6)*(5/6)).
breezeintopl wrote:diogenes wrote:weilan8 wrote:#60 A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is
....
How can we calculate N[(70-mu)/sigma]?just by hand?or we have the distrbution form at the back of our test paper?
mrb wrote:I also wonder this. There is a similar question on one of the other practice tests. I don't really know much probability. I know the general idea behind how one would calculate this, but I have no idea how to do it quick enough for the test and without tables or computers. I have asked this question on another forum, and seen it asked by a couple different people, and have yet to see anyone give a good answer (I have seen several people who actually did know probability say there is no good way to calculate this as a GRE question).
mtey wrote:Maybe the best way is to memorize some quantiles of the normal distribution like Leduc suggests in Cracking the Gre.
mrb wrote:I also wonder this. There is a similar question on one of the other practice tests. I don't really know much probability. I know the general idea behind how one would calculate this, but I have no idea how to do it quick enough for the test and without tables or computers. I have asked this question on another forum, and seen it asked by a couple different people, and have yet to see anyone give a good answer (I have seen several people who actually did know probability say there is no good way to calculate this as a GRE question).
hadimotamedi wrote:For #60:
We have estimated a binomial distribution with a normal distribution. So why the Chebyshev inequality cannot provide rough estimation on the probability of the required range?
Adarsh Raj wrote:Hi,
Could you also shed some light on Q 59.
Thanks
hadimotamedi wrote:Excuse me, for the Chebyshev inequality , I mean what outlined as :
http://mathworld.wolfram.com/ChebyshevInequality.html
It seems that it must provide fair approximation for the mentioned case.Isn't it?
weilan8 wrote:need some questions for the following questions. thanks!
#60 A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is
(A) greater than 0.50
(B) between 0.16 and 0.50
(C) between 0.02 and 0.16
(D) between 0.01 and 0.02
(E) less than 0.01
Answer:C
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