Question

Consider a normal population distribution with the value of σ known. (a) what is the confidence level for the interval x ± 2.81σ/ n ? (round your answer to one decimal place.)

Answer 1

The confidence level would be 99.5%.

When we know the confidence level, we find the z-score by:
Dividing the CL by 100;
Subtracting this result from 1;
Dividing this result by 1;
Subtracting this result by 1.

Algebraically, this looks like:
$1-(\frac{1-(\frac{CL}{100})}{2})$

Using a z-table (http://www.z-table.com), we see that a z-score of 2.81 results in an area under the curve of 0.9975, so we now have
$1-(\frac{1-(\frac{CL}{100})}{2})=0.9975$

To begin solving this, we will subtract 1 from both sides:
$-(\frac{1-(\frac{CL}{100})}{2})=-0.0025 \\ \\(\frac{1-(\frac{CL}{100})}{2})=0.0025$

We will now multiply both sides by 2:
$1-\frac{CL}{100}=0.005$

Subtract 1 from both sides:
$-\frac{CL}{100}=-0.995 \\ \\\frac{CL}{100}=0.995$

Multiply both sides by 100:
CL = 99.5