Question on probability and the application of standard deviation this is a question on probability which is new to me so I'd appreciate it if someone would tell me why my answers are wrong.


Suppose $10\%$ of all wine bottles have corks (instead of screw-tops).
  A restaurant opens $400$ random bottles in a month and counts the
  number that are corked. Use a normal distribution to estimate the
  probability that the restaurant will open $k$ corked bottles.
(a) What are μ and σ in the normal approximation?

The standard deviation is the average, so it would be $40$ corked bottles. σ is just the standard deviation squared, so that'd be $40^2$.

(b) The probability is about .68 that the number of corked bottles
  will fall between a and b. What are a and b?

Since 68% percent falls between one standard deviation, wouldn't this be $80$?

(c) What are a and b for an approximate probability of .95?

This is two standard deviations, so $160$.

(d) What are a and b for an approximate probability of .99?

Three standard deviations, so $240$.

The restaurant wants the head waiter to avoid serving corked bottles
  since they’re more expensive. So they offer her a bonus if fewer than
  the expected number of corked bottles gets opened, provided the
  results are significant at the .01 level. On the other hand they’ll
  fire her if more than the expected number of corked bottles gets
  opened, assuming the results are significant at the .01 level.
(e) How few corked bottles have to be opened for the head waiter to
  get the bonus? How many corked bottles have to opened for the head
  waiter to lose her job? (Assume the estimates above are accurate.)

I don't get this question, wouldn't this just be the opposite of a standard deviation at 99%?
 A: For (a), just think about how ridiculous the answer seems. $40^2=1600$, meaning that samples are an average of 1600 bottles away from the true mean, when there are only 400 in the population. $\sigma$ is not the standard deviation squared, IT IS the standard deviation. Therefore, $\sigma$ is the $variance$ squared, which is given by $\sigma=\sqrt{npq}$, where $n$ is the sample size, $q=1-p$, and $p$ is the probability of a success. 
For (b), you are correct in stating that $0.68$ is within one standard deviation of the mean, but you initial calculation of standard deviation is incorrect. Again, $\sigma=Standard Deviation=\sqrt{npq}$. This means that $a=\mu-\sigma$ and $b=\mu+\sigma$. 
For (c), $.95$ is 2 standard deviations away from the mean, but again, your standard deviation is incorrect. $a=\mu-2\sigma$ and $b=\mu+2\sigma$. 
For (d), same as (b) and (c), but $a=\mu-3\sigma$ and $b=\mu+3\sigma$.
For (e), since .01 is the significance level, having less than the $1st$ percentile (in bottles) opened means she gets a bonus, and having more than the $99th$ percentile gets her fired.
