Question: I have four questions:

The time that it takes to assemble a piece of machinery is well modeled by the normal distribution with mean of 72.9 minutes and standard deviation of 8.55 minutes.

What is the probability that it will take less than an hour to assemble the next piece of machinery? 

What is the probability that it will take between 65 minutes and 75 minutes to assemble the next piece of machinery? 

What is the probability that it will take more than 80 minutes to assemble the next piece of machinery

Find the 16th percentile of the random variable, time taken to assemble. This is the time that is exceeded 84% of the time

I'm not sure how to approach these questions, can someone guide me through them possibly?

Answer: 
You want to know the following : $P(X\leq60)=P(\frac {X-\mu}{\sigma}\leq \frac {60-72.9}{8.55})$.Then you will find (after calculations...) :$P(Z\leq -1.5)$.Using the symmetry of the normal curve, you must find: $0.5- P(0\leq Z\leq 1,5)$.For a better understanding, take a look at the [curve](http://t3.gstatic.com/images?q=tbn%3aANd9GcR3LV6klFOFsBZ7YLe6nqwhn37Cf-pFnbq_XfngaJhLhmqPVNSy). Then you just have to look at the Normal table (look at the interval you found ), which will give you the probability.

For the other questions you will do just the same.Paying attention to the area.Try to draw the curve and the desired probability.It always helps.
Write $X=$time to assemble a piece of machinery. We model $X$~$N(72.9, 8.55^{2})$.
For Q1. $P(X < 60) = P(Z < (60 - 72.9)/8.55)$. Then, obtain this probability either with tables or software.

(Here $Z$ is a random variable with the standard normal distribution).

The other two are approached similarly.
The last one is slightly different. Here, we want to find $t$ such that
$P(X > t) = 0.84$. Then, we know that $P(Z > (t - 72.9)/8.55) = 0.84$. Again, consult tables or software to find the value of (t - 72.9)/8.55 which satisfies this.