Normal Probabilities Using the Z-Table

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Any normal distribution, which is completely described by the values of its mean, μ, and its standard deviation, σ, can be transformed into the so-called standard normal via the transformation

Z=\frac{X-\mu}{\sigma}

The "Z-table" gives probabilities that values of \left.z\right. are found between 0 and given values. These probabilities are represented by areas under the density curve of the standard normal (which is like a continuous histogram). The total area under this curve is 1. Hence, to the right or to the left of 0 is half the data, and so corresponds to a probability or area of 1/2.

Now here's how we generally use the "Z-table":

Probability of values of  to the right of the mean and up to some value, such as 1.3 here.
Probability of values of \left.z\right. to the right of the mean and up to some value, such as 1.3 here.
This one is straight out of the Z-table:

P(0 \le z \le 1.3)=0.4032

Computing probability of values of  on the left side of the mean, from the mean
Computing probability of values of \left.z\right. on the left side of the mean, from the mean
Because of the beautiful symmetry of the normal density curve, we only need to give values to the right of 0; to get answers for values of \left.z\right. to the left of zero, we simply reflect the problem from the left to the right:

P(0 \le z \le |-1.3|)=P(0 \le z \le 1.3)=0.4032

Both left and right probabilities
Both left and right probabilities
Split the problem into two problems, corresponding to the two cases above:

P(-1.7\le z \le 1.3)=
P(0 \le z \le |-1.7|) + P(0 \le z \le 1.3)=
P(0 \le z \le 1.7) + P(0 \le z \le 1.3)=0.4554+0.4032=0.8586

Probability of values of  that correspond to extreme values to the right of the mean.
Probability of values of \left.z\right. that correspond to extreme values to the right of the mean.
You can imagine removing the area in the first problem above from the entire amount of area on the right hand side (which amounts to .5, or half the total area):

P(z > 1.3)=.5-P(0 \le z \le 1.3)=.5-0.4032=.0968

Probability of values of  that correspond to two arbitrary finite values to the right of the mean.
Probability of values of \left.z\right. that correspond to two arbitrary finite values to the right of the mean.
Once again, imagine removing the area between 0 and .5 from the area between 0 and 1.7:

P(.5 \le z \le 1.7)=
P(0 \le z \le 1.7) - P(0 \le z \le .5)=0.4554-0.1915=0.2639

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