A job interview question on flipping a coin I was asked the following question during a job interview:

A coin is flipped 1000 times and 560 times heads show up. Do you think the coin is biased?

What would be your answer?
(I find the question "Quantifying 'survey bias' in reports" related (but it is not answered).)
 A: Call $X$ the number of heads. 
Assume it is not biased. It is the sum of 1000 independent Bernoulli variables with mean $0.5$ and variance $0.5\times 0.5=0.25$. It has mean $500$ and variance $250$. The standard deviation is $\sqrt{250}\approx 16$. 
Intuitively $X$ should be 500 +/- 16.
$X$ can be approximated by a normal distribution (1000 is large enough). The question is : what is probability for a normally distributed variable to have a distance to the mean at least $60/16=3.8$ times the standard deviation. You can find it in this table : https://en.wikipedia.org/wiki/Standard_normal_table
$p=1-2*0.49993=0.00014$
As a conclusion, if the coin is unbiased, the probability of a number of heads as great as 560 is 0.014%. This is quite small. The coin is biased quite certainly.
Or you can use a $\chi^2$ test https://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test that will yield the same conclusion.
A: The interviewer may also have been using this as a way to see how you nuance language around the discussion of statistical results.  Other answers have made it clear, this is a low probability event if the coin is fair.  For many, that may be enough evidence to claim bias. However, depending on how the interviewer worded the question (and the context leading up to the question) they may be looking for you to make the distinction that while the "best" available evidence points to it being bias, there is of course no way to know this with absolute certainty.  
(Although I would be certain enough evidence for me not to let anyone use that coin to decide who gets the dirty job).
A: With a large number of independent Bernoulli trials, the sample proportion has an approximate normal distribution by the Central Limit Theorem. With $\hat{p}= 0.56$ and $se(\hat{p}) = \sqrt{0.56(1-0.56)/1000} \approx 0.015 $. The sample test statistic for the proportion test of the hypothesis of $p=0.5$ corresponding to the fair coin is $Z \approx (0.56-0.50)/0.015 \approx 4$. Using the normal approximation to the sampling distribution of the test statistic under the null hypothesis, the probability of observing 560 or more, or 440 or less heads is very small, less than 0.001 which is very strong evidence of the coin being unfair.
A: I would talk about normal distributions and standard deviations from the mean.
 Draw a nice normal distribution curve on a board.

Then ASK what is the definition of biased; based on the number of standard deviations from the mean.
A: I like the "easy" and "certified" answer that can come from having some basic resources.  Managers aren't going to understand algebra.  You get 5 bullet points and can't say any math at all, but defend your assertion.  I have been required to do this.  If this is your question in a job interview, especially if the person asking the question doesn't have a math degree, then they want to see if you "speak human".   
I would go to this site
http://epitools.ausvet.com.au/content.php?page=CIProportion
I would type in the numbers, and select 'all confidence-interval methods', and hit "submit".
There are good guidelines for which method to use, but they all give a consistent number for the lower interval that does not include 50%.  
A non-biased coin would include 50% in its confidence interval.
I would say "this is made by world-class PhD's in stats, and is a government facing AI in epidemiology", so without any other reason than this, we might still believe its numbers are good.  Also, all the different methods agree.
Comment:
I was asked in my interview "how many marbles do I need to draw from a bowl in order to make a pair, when there are two colors uniformly randomly distributed", and why.  
A: I would say that it would require some simple calculations. Let $X\sim \operatorname{Binomial}(1000, 0.5)$. If the coin is fair it should be quite likely to get 560 heads out of 1000. So we calculate that probability as: $\Pr(X=560)=\binom{1000} {560}0.5^{560}(1-0.5)^{1000-560}\approx0.00002$. Since the probability of getting 560 heads out 1000 flips if the coin is fair is so very small, I find it very likely to be biased.
