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The tables for Z-scores are given in the form of P = Prob(Z < z) for various value of P and z.

Since Prob(Z > z) = 0.93 > 0.5, then by symmetry, z < 0.

So suppose z = -a where a > 0

Now Prob(Z > -a) = 0.93 is the same as Prob(Z < a) = 0.93 [because the standard Normal is symmetric].

therefore, from the tables, a = 1.4758 (approx)

and so z = -1.4758 (approx).

Q: How do you find the number z such that 93 percent of all observations from a standard Normal distribution are greater than z?

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