Question

Find the Z-scores that separatethe middle 81% of the distribution from the area in the tails of the standard normal distribution.(Use a comma to separate answers as needed. Round to two decimal places as needed.)

Answer 1

Step-by-step explanation

First we must find the z-scores that define the area under the curve of the normal distribution that is centered in the middle and is equal to the 81% of the total area under the curve. Basically we have:

Since the area is centered at z=0 the two z-scores that we are looking for have the same module i.e. they only differ in their sign. In order to find them we must use a z table.

In a z table you can find the area between negative infinite and a given z. If the z-scores that we are looking for are -Z and Z then in a z table we can find the area under the curve for the intervals (-∞,-Z) and (-∞,Z).

Since the 81% area described before is centered at the mid of the curve we know that the remaining areas at its left and right are equal and their sum is equivalent to the remaining 19% of the total area. These two areas are the ones in the intervals (-∞,Z) and (Z,∞) and are equal to 0.19/2=0.095. So in a z table we just need to find the z-score for which (-∞,z)=0.095 and that z-score will be -Z. So we look for it in the z table:

So as you can see the z-score is -1.31 so we have -Z=-1.31 and Z=1.31.

Answer

Then the answer is:

The Z-scores are -1.31,1.31.