Question

given a standardized normal distribution, between what two values of z (symmetrically distributed around the mean) will 95.44 percent of all possible z values be contained?
-1.28 and +1.28
-2.00 and +2.00
-1.00 and +1.00

Explanation:

Step1: Recall the properties of the standard normal distribution

In a standard normal distribution \(Z\sim N(0,1)\), the total area under the curve is \(1\). If the values are symmetric about the mean (\(z = 0\)), and we want to find \(z_1\) and \(z_2\) such that \(P(z_1<Z<z_2)=0.9544\). Then the area in the two - tails is \(1 - 0.9544=0.0456\), and the area in each tail is \(\frac{0.0456}{2}=0.0228\).

Step2: Use the standard normal table (z - table)

We look up the \(z\) - value corresponding to an area of \(0.0228\) in the left - tail. Looking up \(A = 0.0228\) in the standard normal table (the cumulative distribution function \(\varPhi(z)=P(Z\leq z)\)).
We know that \(\varPhi(-z)=1-\varPhi(z)\). If \(\varPhi(z_1) = 0.0228\), then looking at the standard normal table:
For \(z=- 2\), \(\varPhi(-2)=\frac{1}{2}(1 - 0.9544)=0.0228\) (since for a standard normal distribution \(P(-z<Z<z)=1 - 2\varPhi(-z)\)). And by symmetry, if \(z_1=-2\), then \(z_2 = 2\)

Answer:

-2.00 and +2.00