SOLUTION: If the random variable z is a Standard Normal Score, what is P(-2.00 ≤ z ≤ +2.00)? How did you find this probability?

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Question 1140431: If the random variable z is a Standard Normal Score,
what is P(-2.00 ≤ z ≤ +2.00)? How did you find this
probability?

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
From the cumulative normal distribution table,
P%28z+%3C=+2.00%29+=+0.97725.
P%28z+%3C+-2.00%29
= 1+-+P%28z+%3E=+-2.00%29
= 1+-+P%28z+%3C=+2.00%29+.....by symmetry of the normal distribution
= 1+-+0.97725
= 0.02275

P%28-2.00+%3C=+z+%3C=+2.00%29
= P%28z+%3C=+2.00%29+-+P%28z+%3C+-2.00%29
= 0.97725+-+0.02275
= 0.9545+->answer
P%28-2.00+%3C=+z+%3C=+%2B2.00%29=0.9545+