Table for a standard normal distribution of z z0 z1 z128 z16

Table for a standard normal distribution of z

z=0

z=1

z=1.28

z=1.645

z=1.96

z=2.33

P(Z<=0)=0.5

P(Z<=1)=0.84

P(Z<=1.28)=0.9

P(Z<=1.645)=0.95

P(Z<=1.96)=0.975

P(Z<=2.33)=0.99

Given above table, the probability P(Z>=2.33) is

Solution

we have

P( z< 2.33) = 0.99

we know that

P( z > 2.33 ) = 1 - P( z < 2.33)

so the answer is

P ( z > 2.33) = 1 - 0.99

P( z > 2.33) = 0.01