How many batteries will be working after 280 minutes? 
A study of data collected at a company manufacturing flashlight
  batteries shows that a batch of 8000 batteries have a mean life of 250
  minutes with a standard deviation of 20 minutes. Assuming a Normal
  Distribution, estimate:
How many batteries will continue working after 285 minutes? 

This is my answer to this question does it look correct or are there any improvements I can make?
Batch: 8000
Mean: 250 minutes 
SD: 20 minutes


(285-250)/20 = 1.75
Z-Score of 1.75 = .4599
(.5-4599)*8000 = 320.8 batteries will be working 

 A: Using R software in which pnorm is a normal CDF.
Lifetime $X$ of any one battery has $X  \sim \mathsf{Norm}(\mu = 250,\sigma=20).$ As you say, you need $8000\,P(X>285).$
8000*(1 - pnorm(285, 250, 20))
[1] 320.4733

Round to integer 320.
Standardizing and using printed tables of the standard normal CDF.
$$P(X > 285) = P\left(\frac{X-\mu}{\sigma} >
\frac{285 - 250}{20}\right)\\ = P(Z > 1.75) \approx 0.0400.$$
Then .04(8000) = 320.
Note: Using R you get more decimal places of accuracy than the usual four place accuracy of printed tables.
I think the problem has been 'rigged' to give you an
integer answer from printed tables.
You have the right idea, but you could give a little clearer explanation of your method.
A: I think you have the right idea, but perhaps your presentation could be improved.
Let $X$ be the random variable denoting the life of a battery, in minutes.
$\because X$ follows a normal distribution
$\therefore X \sim N(250,20^2)$
Find $P(X > 285)$, that is, the probability of a single battery lasting more than 285 minutes.
Can you take it from here?
