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I got to calculate the value for the variable $c$ when they give me this intervals but I don't know how to interpret the inequations.

a) $P(-c \le z \le c) = .668$.

b) $P(c \le |z|) = .016$.

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    You are right I just was wrong on the question title. It's about normal distribution.2012-10-08

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So. We have a probability variable $z$: some random thing happens, and then $z$ gets a value, assumed as a real number.

If this 'random thing' occurs according to a given distribution, then the corresponding distribution function is defined as: $F(t):=P(z If the distribution is continuous, then all single points have zero probability, so $F(t)=P(z\le t)$. And, (assumed that $z$ has value with $1$ probability:) we also have $P(z>t) = 1-P(z\le t)$.

In your case, if you have given a continuous distriution by $F$, then $\text{a) }\ P(-c\le z\le c) = F(c)-F(-c)$ $\text{b) }\ P(c\le|z|) = P((z\ge c)\lor(z\le -c))=1-F(c) + F(-c) $

Now, if it happens to be the standard normal distribution (that is, in a sense a limit of binomial distributions), then $F=\Phi$ and you can also use its symmetry: $\Phi(-c) = 1-\Phi(c)$.