Question

complete the statement. round your answers to the nearest thousandth. in a population that is normally distributed with mean 99 and standard deviation 4, the outside 88% of the values are those less than or greater than.

Explanation:

Step1: Find the z - score

Since we want the outside 88% of the values, the middle 12% of the values. So the area to the left of the lower z - score is \((1 - 0.88)/2=0.06\).
Using a standard normal table or calculator (inverse - normal function), the z - score \(z_1\) corresponding to an area of \(0.06\) is approximately \(z_1=- 1.555\).
The z - score \(z_2\) corresponding to an area of \(1 - 0.06 = 0.94\) is approximately \(z_2 = 1.555\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 99\), \(\sigma = 4\), and \(z\) is the z - score.
For \(z=-1.555\):

$$
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For \(z = 1.555\):

$$
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Answer:

The outside 88% of the values are those less than \(92.780\) or greater than \(105.220\)