0
$\begingroup$

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?

  • 0
    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

1 Answers 1

1

Raskolnikov has essentially given the answer, but to spell it out for other readers:

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 $