Question

find the z - scores that separate the middle 77% 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 curve is \(1\). If the middle area is \(0.77\), then the area in the two tails is \(1 - 0.77=0.23\). The area in each tail is \(\frac{0.23}{2}=0.115\).

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

We want to find the \(z\) - score \(z_1\) such that \(P(Z\lt z_1)=0.115\). 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 \(z_1\approx - 1.20\) (using a calculator: \(z=\text{invNorm}(0.115)\)).

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

Since the standard normal distribution is symmetric about \(z = 0\), if \(z_1\) is the \(z\) - score for the left - tail, then the \(z\) - score for the right - tail \(z_2=-z_1\). So \(z_2\approx1.20\)

Answer:

\(-1.20,1.20\)