Question

Find the value of z /2 that corresponds to a confidence level of 89.48%.

Answer 1

Given that 
P(z/2)=89.48%
the value that corresponds to this will be given as follows:
the z-value that corresponds to 89.48% in the z-table is:
1.25
thus
z/2=1.25
hence
z=1.25*2
z=2.50

Answer 2

Answer:  $1.6202$

Step-by-step explanation:

The given confidence level : c= 89.48%.

⇒ c= 0.8948   [ in decimal]

Then, the significance level = $\alpha= 1-c$

$=\alpha= 1-0.8948=0.1052$

Now, to find the value of $z_{\alpha/2}$ , we need to find the two -tailed z-value using z-table for $\alpha=0.1052$ or we can also find the one-tailed z-value for  for $\alpha/2=0.1052/2=0.0526$

Using z-value table , the value is $1.6202$

Hence , $z_{\alpha/2}=1.6202$ that corresponds to a confidence level of 89.48%.