Question

which z-values correspond to the outside 16% of the standard normal distribution?
round your answers to the nearest thousandth.
z <
or
< z

Explanation:

Step1: Find the area in one tail

Since the outside area is \(16\%\) or \(0.16\), the area in one tail is \(\frac{0.16}{2}=0.08\).

Step2: Use the standard - normal table (or inverse - normal function)

We want to find \(z\) such that \(P(Z < z)=0.08\) and \(P(Z>z) = 0.08\) (by symmetry of the standard - normal distribution).
Using the inverse - normal function (e.g., on a calculator, if \(X\sim N(0,1)\), and we use the formula \(z=\text{invNorm}(p)\) where \(p\) is the cumulative probability).
For \(p = 0.08\), \(z\approx - 1.405\) (using a calculator: \(\text{invNorm}(0.08)\approx - 1.405\)).
By symmetry, the other \(z\) - value is \(z = 1.405\) (since the standard - normal distribution \(y = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}\) is symmetric about \(z = 0\), i.e., \(P(Z < -a)=P(Z>a)\)).

Answer:

\(z<-1.405\) or \(1.405 < z\)