Hypothesis testing: carrying capacity of bridges 
A manufacturer claims that, at the moment of stockage, the carrying
  capacity (CC) of the produced bridges follows a normal distribution
  with $\mu = 2500\, kg$ and $\sigma^2=150^2\, kg^2$. Moreover the CC of
  the different bridges are independent. The CC of the bridges decreases
  as time goes by, so after 2 years the manufacturer has to decide
  whether to get rid of the bridges or not (independence, normality and
  variance are assumed to be preserved!). 
How many bridges do we have to test such that 
  
  
*
  
*if the average CC is $2500\, kg$, then this can be claimed with a certainty of $95\%$.
  
*if the average CC decreases by $150\, kg$, then it is possible to detect this decrement with a probability of $90\%$.

My attempt:
We set $H_0: \mu = 2500$ and $H_A: \mu < 2500$. Suppose that $\bar{X}$ denotes the average CC of bridges in a sample of size $n$. Then it is clear that $\bar{X} = N(2500,150^2/n)$ under $H_0$, where $n$ is the unknown we're solving for. 
The first bullet points assumes that we're working under the assumption that $H_0$ is true. The second one states that $\bar{X} = N(2350,150^2/n)$. I feel like I'll have a system of two equations to solve for $n$. I just don't know how the express 'being able to detect with a probability of $\alpha\%$' mathematically. I feel like confidence intervals are an option. 
For the first one, we have that the sample average will lie in $$[2500-1,96\cdot \frac{150}{\sqrt n}, 2500+1,96\cdot \frac{150}{\sqrt n}].$$ And then we can do the same for the second bullet point. But how do I find $n$ from these intervals?
Thanks.
 A: It seems to me there is some missing info in the first case, though I am not really sure. Rejection/acceptance criterion is however given below. $Z = \frac{\bar{X} -2500}{150/\sqrt{n}}$. 
The second one I believe we are able to solve as following: 
Having a rejection criterion from the hypothesis test $Z < k$, where $Z$ is your test observator, standard normally distributed under $H_0$. Let's say you have one of 95%, then $k = -1.64$ (remember to use in this case a one sided interval!). Now, I guess what they are asking for, is finding the probability of rejecting $H_0$ to be 90%, i.e. 
$$
P(Z<k \;|\; \mu = 2350) = 0.9, \quad Z = \frac{\bar{X}-2500}{150/\sqrt{n}}, \quad k = -1.64
$$
In this case, $Z$ is no longer standard normally distributed, hence we have to manipulate the inequality, by adding $150/(150\sqrt{n})$. Then the left hand side is normally distributed, thus
$$
P(Z^* < -1.64 + \sqrt{n}) = 0.9 \quad \Rightarrow -1.64 + \sqrt{n} = 1.28 \; \text{(from quantile table)}
$$
From this, $n$ has to be at least $8.52$, thus this answer will give $n=9$. 
