Quick Probability Question (Exam Prep): Please Help? 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
the answer is C.
Why is that? I know the probabily = P(a six comes up exactly 70 times) + P(a six comes up exactly 71 times) + ... + P(a six comes up exactly 360 times),
which is a sum of binomial probabilities, but how do we approximate it?
Thank you very much.
 A: The probability of rolling a $6$ is $1/6$, and the complementary probability is $5/6$, so the expected number of $6$’s is $(1/6)\cdot 360=60$, with standard deviation $\sqrt{360(1/6)(5/6)}=\sqrt{50}\approx 7$. The distribution is approximately normal, so you can estimate the desired probability as the probability that a normally distributed random variable is at least $(70-60)/7=10/7$ standard deviations above the mean. The $68-95-99.7$ rule of thumb tells you that this is somewhere between $\frac12(1-0.68)=0.16$ and $\frac12(1-0.95)=0.025$, and C is the only answer that fits, even without using any accurate calculations.
By the way, dice is plural: the singular is die.
A: Use the central limit theorem and a continuity correction.
The expected number of $6$s is $360/6=60$.  The standard deviation of the number of $6$s is $\sqrt{360\cdot(1/6)\cdot(5/6)}= \sqrt{50} = 5\sqrt{2}$.
"70 or more" is the same as "more than 69".  That's where the "continuity correction" tells you to use $69.5$ when you use a continuous approximation to this discrete distribution.
So $$\frac{69.5-\text{expected value}}{\text{standard deviation}} = \frac{69.5-60}{5\sqrt{2}} \approx 1.3435.$$ You're asking: what's the probability of being more than that many standard deviations above the mean.  Plug that number into the appropriate table or software for the normal distribution.
BTW, I'd have said "a fair die", and used "dice" only as the plural.
Later edit: I find it being pointed out that calculators aren't allowed, as if that means this is all of no use.  But the answer was phrased as "between" something and something.  So you shouldn't need a calculator to apply this; you don't need a precise number, but only that it's between something and something else.  You don't need a calculator to know immediately that $\sqrt{50}$ is slightly more than $7$, so that $9.5/\sqrt{50}$ is somewhat more than $1$.  Even before doing any of this you know that the probability is less than $0.5$, since $70$ is well above the expected value.  And "somewhat more than $1$ is certainly less than $2$.  In this case, all you really need is that it's between $1$ and $2$ standard deviations above the mean.  So what you see above should do it even if you don't have a calculator.
