Finding probabilities involving Z scores

Example 2AFind the area to the left of a negative Z scoreFind P( z < – 1.02 )The area under the z curve that is to the left of a given z value represents the probability ofselecting one number from the z distribution and having that number be less than the z value.If we want to find the probability of selecting one number from the z distribution and having thatnumber be less than the z value of – 1.02 we need to find the area to the LEFT of z = – 1.02The number at the intersection of the –1.0 row and the 0.02 column is 0.1539Negative Z ScoresStandard Normal (Z) Distribution: Cumulative Area to the LEFT of ZZ 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09–1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379This means that the area to the left of z = –1.02 is 0.1539the area to the left ofz = – 1.02 is .1539left tail area= .1539z = – 1.02negativevalue0If the area to the left of z = – 1.86 is .1539 thenP( z < – 1.02 ) = 0.1539Section 6 – 3B Lecture Page 5 of 16 © 2012 Eitel