Part 1 Use the standard normal z score table to find P100 z

Part 1

Use the standard normal (z score) table to find P(-1.00 z):

Part 2

Find the probability that a data value picked at random from a normal population will have a standard score (z) that lies between the following pairs of z-values. z = 0 to z = 2.10:

Solution

part 1

P( z > -1.00) = 1 - 0.15870.8413

part 2

P( 0 < z <2.10 ) = P( z > 0 ) - P( z > 2.10 ) = 0.50 - 0.0179 = 0.4821