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The monthly starting salaries of students who receive an MBA degree have a standard deviation of $110. What size sample should be selected to obtain a 0.95 probability of estimating the mean monthly income within $20 or less

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Answer:

116.21

Explanation:

Using the formula for calculating margin error to tackle the question.

Margin error =
(Z \sigma)/(√(n) )

Z is the value at 95% confidence


\sigma 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
\sigma = $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;


M.E = (Z \sigma)/(√(n) )\\\\√(n) = (Z \sigma)/(M.E ) \\\\


n = ((Z \sigma)/(M.E ))^(2)

Substituting the give value into the resulting expression;


n = ((1.96 * 110)/(20))^(2)\\\\n = ((215.6)/(20)) ^2\\\\n = 10.78^2\\\\n = 116.21

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

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