Question

find the z - scores that separate the middle 67% of the distribution from the area in the tails of the standard normal distribution.
the z - scores are □.
(use a comma to separate answers as needed. round to two decimal places as needed.)

Explanation:

Step1: Calculate the area in the tails

The middle area is \(67\% = 0.67\). The area in the two tails is \(1 - 0.67=0.33\). The area in each tail is \(\frac{0.33}{2}=0.165\).

Step2: Find the z - score

We want to find the \(z\) - score \(z\) such that \(P(Z < z)=0.165\) (left - tail) and \(P(Z>z) = 0.165\) (right - tail). Using the standard normal table or a calculator with a normal - distribution function (e.g., in Excel, using the function NORM.S.INV(0.165)), we find the \(z\) - score.
For the left - tail, \(z=- 0.98\) (using the standard normal table: looking for the value closest to \(0.165\) in the body of the table). Since the standard normal distribution is symmetric, the right - tail \(z\) - score is \(z = 0.98\).

Answer:

\(-0.98,0.98\)