Question

1. what proportion of a normal distribution is located between each of the following z - score boundaries? a. z = - 1.64 and z = + 1.64

Explanation:

Step1: Recall the properties of the standard - normal distribution

The standard - normal distribution has a mean of 0 and a standard deviation of 1. The total area under the curve is 1. We want to find \(P(-1.64<Z<1.64)\).
We know that the cumulative - distribution function of the standard - normal distribution is denoted as \(\varPhi(z)\), and \(P(-1.64<Z<1.64)=\varPhi(1.64)-\varPhi(-1.64)\).

Step2: Use the symmetry of the standard - normal distribution

The standard - normal distribution is symmetric about \(z = 0\), so \(\varPhi(-z)=1 - \varPhi(z)\). Then \(\varPhi(-1.64)=1-\varPhi(1.64)\).
So \(P(-1.64<Z<1.64)=\varPhi(1.64)-(1 - \varPhi(1.64))\).

Step3: Look up the value of \(\varPhi(1.64)\) in the standard - normal table

From the standard - normal table, \(\varPhi(1.64)=0.9495\).
Then \(P(-1.64<Z<1.64)=0.9495-(1 - 0.9495)=0.9495 - 0.0505=0.899\).

Answer:

0.899