(1) area above a certain value

(2) area below a certain value

(3) area between two values

(4) area outside of two values

Then, draw a picture of the normal curve, with the mean in the center (representing a standard score of z=0), and the other applicable values in their approximate locations.

Now, convert the scores in your question into z scores:

z = (x - u) / s

Usually, for the normal probability table (or z table) in the back of a statistics book, the table gives you the percentage of the area between negative infinity and z.

If the question asks for (1), you find the area below z and subtract it from 1 (since the area above a given value of z and the area below that value of z must =1).

For (2), just look up the area below z.

For (3), find the area below the larger z, then subtract the area below the smaller z. What you're left with is the area in between the z values.

For (4), find the area below the larger z and subtract it from 1, then find the area below the smaller z, then add it to the area beyond the larger z.

As soon as I get a chance, I'll add some real examples here.

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Jin ("quark") has provided some examples here:

P(Z< -0.5) - P(Z< -2)

=[1-P(Z<0.5)] - [1-P(Z<2)]

=[1-0.6915]-[1-0.9772]

=0.2857

P(Z> -0.5)

=P(Z<0.5)

=0.6915

P(Z>=5)

=1-P(Z<=5)

=1-0.000

=1 approx.

P(Z< -2.18)

=1-P(Z<2.18)

=1-0.9854

=0.0146

P(1 < Z < 2.5)

=P(Z<2.5) - P(Z<1)

=0.9938-0.8413

=0.1525

P(Z>=2.5)

=1-P(Z<=2.5)

=1-0.9938

=0.0062

P(-2 < Z < 2.5)

=P(Z<2.5) - P(Z< -2)

=P(Z<2.5) - [1-P(Z<2)]

=0.9938-[1-0.9772]

=0.9710

P(Z>1.43)

=1-P(Z<1.43)

=1-0.9236

=0.0764