find the value za/2 that corresponds to a confidence level 98.22%

Question

I need it asap

Answer 1

since the confidence level is 0.9822, α= 1-0.9822 = 0.0178. Thus, α/2 = 0.0089. Thus, the standard normal variate that is exceeded by only 0.0089 of the standard normal variates is z_0.0089 =2.369752.