Answer:
Step-by-step explanation:
Using the formula for calculating margin error to tackle the question.
Margin error = [tex]\frac{Z \sigma}{\sqrt{n} }[/tex]
Z is the value at 95% confidence
[tex]\sigma[/tex] is the standard deviation
n is the sample size to be estimated
Since the mean monthly income is within $20 or less, our margin error will be $20
Given [tex]\sigma[/tex] = $110, Z value at 95% confidence = 1.96 we can calculate the sample side n.
Making n the subject of the formula from the equation above;
[tex]M.E = \frac{Z \sigma}{\sqrt{n} }\\\\\sqrt{n} = \frac{Z \sigma}{M.E } \\\\[/tex]
[tex]n = (\frac{Z \sigma}{M.E })^{2}[/tex]
Substituting the give value into the resulting expression;
[tex]n = (\frac{1.96 * 110}{20})^{2}\\\\n = (\frac{215.6}{20}) ^2\\\\n = 10.78^2\\\\n = 116.21[/tex]
This shows that the sample size that should be selected to obtain a 0.95 probability of estimating the mean monthly income within $20 or less is approximately 116.21
