# A normal distribution question is this the correct way of going about it?

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:

(i) How many batteries will fail before 220 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

(250-220)/20 = 1.5

Z-score of 1.5 = .4332

(.5-.4332)8000 = 534.4 will fail after 220 minutes


Let $$X$$ be the random lifetime of a battery. We are told that $$X \sim \operatorname{Normal}(\mu = 250, \sigma = 20),$$ that is to say, a battery selected at random will have a lifetime that is normally distributed with mean lifetime $$\mu = 250$$ minutes, and standard deviation $$\sigma = 20$$ minutes.
The probability that a single randomly selected battery will have lifetime less than $$220$$ minutes is $$p = \Pr[X \le 220] = \Pr\left[\frac{X - \mu}{\sigma} \le \frac{220 - 250}{20}\right] = \Pr[Z \le -1.5] \approx 0.0668072,$$ where $$Z \sim \operatorname{Normal}(0,1)$$ is a standard normal random variable. This means any single battery has only about a $$6.68\%$$ chance of not lasting more than $$220$$ minutes.
Consequently, on average, out of $$n = 8000$$ batteries, we would expect $$np = (8000)(0.0668072) = 534.458$$ batteries in the batch to fail within $$220$$ minutes. That is not to say exactly this many batteries will fail by then, because this number is itself a random variable that follows a binomial distribution with parameters $$n = 8000$$ and $$p = 0.0668072$$.