Confusion - Am I on the right track on this power computation? So the question asks: It has been generally believed that average daily caffeine consumption, $μ$, is at most $200$ mg with $σ^2 = 225$.  To examine whether at present time $μ$ is greater than $200$ mg, it is agreed that daily caffeine consumption of a random sample of $100$ adults be examined.  If the sample mean is greater than $203.50$, the decision will be to reject the null hypothesis. 
The following information may be useful: pnorm(0.67)$=0.75$, pnorm(1)$=0.84,$ pnorm(1.56)$=0.94$, pnorm(2.33)$=0.99$
(a) What are the null and alternative hypotheses?
(b) What is the test statistic and what is it distribution?
(c) What is the probability of a Type 1 error?
(d) What is the power of the test if the true mean is 205mg?
(e) Is the assumption that the population distribution of daily caffeine consumption is normal necessary
for the hypothesis test? Explain your reasoning.
My answers: 
(a) $H_0: μ \le 200$ (average daily caffeine consumption is at most $200mg$)
$H_1: μ > 200$ (average daily caffeine consumption is more than $200mg$)
(b) For large sample, and known population standard deviation, use 1-sample $Z$ test. Distribution: Normal.
$$Z=\frac{\bar X-μ}{\sigma/\sqrt{n}}
=\frac{203.50-200}{15/10}
=2.3333.$$
(c) probability of Type I error, $\alpha=0.05$
(d) Since it is a right-tailed test, therefore, one commits a Type II error (fail to reject $H_0$), when one gets a test statistic less than 1.645. Compute $X_\text{critical}$ by substituting the value in following Z score formula
$Z =1.64=\frac{X_\text{critical}-\mu}{\sigma/\sqrt{n}}=\frac{X_\text{critical}-200}{1.5}$
$X_\text{critical}=202.46$
$$P(\bar X<202.46)=P\left(Z<\frac{202.46-205}{1.5}\right)=P(Z<-1.69333333)=0.0455.$$
Power of test: $1-0.0455 =0.9545.$
(e) Not necessary because the sample size is sufficiently large. 
So I am not sure about my answers in d and e. Should I calculate the new x, then use z-test to get the power of the test? And in e, does my answer sound reasonable enough?
 A: (c) To check your power computation, here is a printout from Minitab's 'Power and
Sample Size' procedure. [The slight discrepanacy might be because you have
used 1.64 instead of 1.644854 (or 1.645 as in many tables).]
 1-Sample Z Test

 Testing mean = null (versus > null)
 Calculating power for mean = null + difference
 α = 0.05  Assumed standard deviation = 15

             Sample
 Difference    Size     Power
          5     100  0.954340

The power curve shows the power against alternatives with
differences other than $\Delta = \mu_a - \mu_0 = 205 - 200 = 5.$
[The only point on this curve of 'P(Reject) vs. $\Delta$' that is
not a 'power' value is above $\Delta = 0$ where the height of
the curve is the significance level $\alpha = .05.$]

(d) My answer would be only slightly more cautious: Because of the large
sample size, probably not necessary--unless the actual population distribution
is extremely skewed or otherwise very far from normal. 
Even for $n=100,$ results would not
be accurate for an exponential population, and would be totally wrong
for a Cauchy population. (But I think the
question clearly expects you to take the Central Limit Theeorem into
consideration.)
