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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 
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Your answer is incorrect.

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 $p = 0.0668072$.

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