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I'm looking at a textbook problem, which has the answer as well, but don't understand what's going on.

The problem: A part with lifetime X has a normal distribution with mean of 7 years and standard deviation of 1.2 years. If the warranty is for 5 years, what % will fail inside the warranty?

The answer is: P(X<=5) = P(Z<=(5-7)/1.2) = P(Z<=-1.67) = .5-.4525 = 0.0475

I follow everything until the .5-.4525 part. Where do these numbers come from?

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    I suppose your textbook is working with some tables for standard normal variables. They contain values for $\mathbb{P}(0. Using these values, you can express the sought after probability.2011-03-09
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    Ah, I've been looking at a different table (the book doesn't have the z-score table). This explains a lot. Many thanks @Raskolnikov!2011-03-09
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    You should also be aware that different books use different tables: some books will have tables for $P(Z \le z)$, some from $P(0 \le Z \le z)$, and at least one common statistics textbook has a table of $P(-z \le Z \le z)$.2011-03-10

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