# Normal Distribution

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 $\mathbb{P}(0<Z<z)$. Using these values, you can express the sought after probability. –  Raskolnikov Mar 9 '11 at 23:43
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! –  user8045 Mar 9 '11 at 23:55
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 $P(-z \le Z \le z)$. –  Michael Lugo Mar 10 '11 at 0:10

For a random variable $Z$ with a standard normal distribution with mean 0 and variance 1, $$Pr(Z\le -1.67) = Pr(Z \ge 1.67) =Pr(Z \ge 0) - Pr(0\le Z < 1.67) \approx 0.5 - 0.4525$$