How many batteries will be working after 280 minutes?

Question

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 

Answer 1

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.

Answer 2

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?