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More than once I have seen this sort of frustrating question on a Mathematics GRE practice test:
A fair die is tossed 360 times. The probability that a six comes up on 70 or more tosses is...
a) greater than .50
b) between .16 and .50
c) between .02 and .16
d) between .01 and .02
e) less than .01

I think the question wants me to approximate the binomial distribution with a normal distribution centered around $360/6 = 60$ with standard devation $\sqrt{360*1/6*5/6} = 17.3...$ Even with this approximation, how would I mentally (or quickly using pen and paper) be able to estimate the CDF to the degree of accuracy that this question asks for?

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1 Answer 1

up vote 2 down vote accepted

The standard deviation is $\sqrt{50}$, a little bit over $7$. So $70$ is $1.4$ standard deviation units up from the mean.

If one recalls some calculations with tables of the standard normal, one may remember that the probability that $Z\gt 1$ is about $0.16$, and the probability that $Z\gt 1.645$ is about $0.05$. That is enough information to make the right choice among those offered. Even if we do not remember about $1.645$, the more commonly occurring $1.96$ ($95\%$ confidence interval) shows that the tail probability cannot, in our case, be less than $0.025$.

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Wow so the test makers contrived to make a question to test my knowledge of specific values of the normal distribution? I will be sure to memorize the values that you mentioned. Thanks! –  Mark Mar 12 '13 at 7:54
    
I do not think this is a particularly good question, except in a Stats subject exam, where one can be confident that there has been sustained exposure. One can of course immediately eliminate (a), and probably (d) and (e), but one bumps into $1$ standard deviation unit less often, since it does not come up in standard estimation problems. –  André Nicolas Mar 12 '13 at 8:07
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