Question

find the critical value ( z_{alpha / 2} ) that corresponds to the confidence level 82%.
( z_{alpha / 2}=square )
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the significance level $\alpha$

The confidence level is $C = 82\%=0.82$. Using the formula $\alpha=1 - C$, we have $\alpha=1 - 0.82 = 0.18$.

Step2: Calculate $\frac{\alpha}{2}$

$\frac{\alpha}{2}=\frac{0.18}{2}=0.09$.

Step3: Find the $z -$score

We want to find $z_{\alpha/2}$ such that $P(Z>z_{\alpha/2})=\frac{\alpha}{2}=0.09$. Then $P(Z\leq z_{\alpha/2})=1 - 0.09 = 0.91$.
Looking up in the standard normal distribution table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: invNorm(0.91,0,1)), we find that $z_{\alpha/2}\approx1.34$.

Answer:

$1.34$