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
| 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 |
Solution
P( Z >2.33) = 1 - P(Z < 2.33)
P( Z> 2.33 ) = 1-0.99
P( Z > 2.33) = 0.01
