Question

find the z - scores that separate the middle 72% 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 total area under the standard normal distribution curve is \(1\). If the middle area is \(0.72\), then the area in the two tails is \(1 - 0.72=0.28\). The area in each tail is \(\frac{0.28}{2}=0.14\).

Step2: Find the \(z\) - score for the left - tail

We want to find \(z\) such that \(P(Z\lt z)=0.14\). Using a standard normal table (or a calculator with a normal - distribution function, e.g., the inverse of the cumulative distribution function for the standard normal distribution \(\Phi^{-1}(p)\)), we find the \(z\) - score.
Looking up \(p = 0.14\) in the standard normal table, \(z\approx - 1.08\)

Step3: Find the \(z\) - score for the right - tail

Due to the symmetry of the standard normal distribution (\(Z\sim N(0,1)\)), if the left - tail \(z\) - score is \(z_1\) and the right - tail \(z\) - score is \(z_2\), and \(P(Z\lt z_1)=0.14\), then \(P(Z\gt z_2)=0.14\) and \(z_2=-z_1\). So \(z_2 = 1.08\)

Answer:

\(-1.08,1.08\)