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Since this is a frequently encountered problem, the answers have been set out in the tables of the standard Normal distribution. We can simply look up the solution. However, since there is an infinite number of Normal distributions (for every combination of 1 and o2) it would be an impossible task to tabulate them all. The standard Normal distribution, which has a mean of zero and variance of one, is therefore used to represent all Normal distributions. Before the table can be consulted, therefore, the data have to be transformed so that they accord with the standard Normal distribution.

The required transformation is the z score, which was introduced in Chapter 1. This measures the distance between the value of interest (180) and the mean, measured in terms of standard deviations. Therefore we calculate and z is a Normally distributed random variable with mean 0 and variance 1, i.e. z ~ N(0, 1). This transformation shifts the original distribution ¡i units to the left and then adjusts the dispersion by dividing through by o, resulting in a mean of 0 and variance 1. z is Normally distributed because x is Normally distributed. The transformation in (3.8) retains the Normal distribution shape, despite the changes to mean and variance. If x had some other distribution then z would not be Normal either.

It is easy to verify the mean and variance of z using the rules for E and V operators encountered in Chapter 1:

Evaluating the z score from our data we obtain

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