Probability

[vvaann]

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 0.5
B. between 0.16 and 0.5
C. between 0.02 and 0.16
D. between 0.01 and 0.02
E. less than 0.01

Answer key:

SPOILER: C

Br,
vvaann

[matroid]

I have some kind of "solution" but I don't like it. If anybody has a better solution, please post it!

(And another disclaimer: I've never learnt or read probability/statistics in English, but maybe you'll understand...)

So, what we have is a binomial distribution B(N, p) with parameters N=360 and p=1/6. We can approximate this by a normal distribution N(M, d) with the same mean and deviation:

M = Np = 60,
d = sqrt(Npq) = sqrt(50) ~ 7.

We have to calculate the probability Prob( N(60, 7) > 70 ) ~ Prob( N(M, d) > 1.5d ). If you are a statistics pro (unlike me), you may know by heart that approximately 93.3% percentile belongs to 1.5 deviation. (I looked it up in my probability book...) Which means that the probability we are looking for is approximately 6.7%, so the answer is what you have given in the spoiler.

Please, please, please, give a nicer explanation,

Md.

PS: I checked it with a computer (unfortunately I failed to smuggle one into the test center ). It says the correct result is 9.14%, which is also C, but it shows that the estimations above are not really good and we've been a little lucky to get the correct answer!

[vvaann]

I also hate this kind of question.

I posted it because I was thinking that I didn't know the 'standard' solution which should be well reasoned and easy-to-understand. If there is no such a solution, I will definitely give up analyzing the question further.