You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ = 54.5 . You would like to be 99% confident that your estimate is within 5 of the true population mean. How large of a sample size is required

  (3)

So the answer for this case would be n=791 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

Since the Confidence is 0.99 or 99%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that

The margin of error is given by this formula:

   (2)

And on this case we have that ME =+5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

  (3)

So the answer for this case would be n=791 rounded up to the nearest integer

step-by-step explanation:

find the vertical and horizontal asymptotes of the function f x = 5 x − 1 . to find the vertical asymptote, equate the denominator to zero and solve for . since the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote is .

step-by-step explanation: