Finding normal probabilities (battery life) The lifespan of a calculator battery is normally distributed with a mean of 1100 days & standard dev. of 60 days. $$\\$$
1) What percent of batteries is expected to survive more than 1200 days?
2) What percent of batteries will survive fewer than 800 days?
3) What length of warranty is needed so that no more than 10% of the batteries will be expected to fail during the warranty period? $$\\$$
This is what I have so far:
1) $P(x > 1200)$
= $1-P(\frac{1200-1100}{60})$
= $1-P(1.67)$
= $1-0.9525$
= 0.0475
2) $P(x < 800)$
= $P(\frac{800-1100}{60})$
= $P(-5)$
= 0
I'm not sure how to do #3.
I would appreciate your help, thanks!
 A: Perhaps I can be of the most help by encouraging you to use notation
that makes sense.  Your write:
$$P(x > 1200) = 1-P\left(\frac{1200-1100}{60}\right) = 1-P(1.67) = 1-0.9525 = 0.0475,$$
in which the second and third terms contain no events and so make no sense.
I suggest something like this: You have 
$X \sim \mathsf{Norm}(\mu = 1100,\, \sigma=60)$ and seek
$P(X > 1200).$ Then
$$P(X > 1200) = 1 - P(X \le 1200) = 
1-P\left(\frac{X-\mu}{\sigma} \le \frac{1200 - 1100}{60}\right)\\
= 1 - P(Z \le 1.67) = 1 -  0.9525 = 0.0475,$$
where the approximate numerical answer can be obtained from printed tables of the standard
normal CDF.
You can get a slightly more accurate answer using software without having to standardize. For
example in R statistical software. The improved accuracy is because rounding
is avoided. (In R, pnorm is the CDF of the normal distribution with mean
and SD given in the second and third arguments.)
1 - pnorm(1200, 1100, 60)
## 0.04779035


Then in the last part, you seek $w$ such that $P(X \le w) = 0.10.$ So write
$$P(X \le w) = P\left(\frac{X - \mu}{\sigma} \le \frac{w-1100}{60}\right)
= P(Z \le (w-1100)/60) = 0.10,$$
Then by normal tables $(w-1100)/60 \approx -1.28,$ and you can solve for $w.$
A slightly more accurate answer can be obtained by software:
 qnorm(.1, 1100, 60)   # 'qnorm' is the inverse CDF or 'quantile' function
 ## 1023.107
 pnorm(1023, 1100, 60) # as a check
 ## 0.09968766

Notice (as some Commenters did not) that the company offering the warranty is stingy, not
wanting to pay out for more than 10% of purchases.

Also, it usually helps to make sketches. Here is a graph of the density
function of $\mathsf{Norm}(1100, 60),$ with vertical red lines marking
locations of interest in the parts above.

Of course, you can't sketch with such good accuracy just by hand, but with
a little effort you can learn to make a facsimile of a normal
density curve that is a lot better than no sketch at all.
