Answer:
The sample size should be at least 79.
Step-by-step explanation:
Let Y be be the resting glocose level of a member of the population chosen randomly, then Y is a random variable with unkown mean λ and Standard deviation σ = 27.
Let X be the sample mean of a sample of lenght n. X has the same mean as Y and the standard deviation is
If n is reasonable high, the Central Limit Theorem states that X has distribution approximately Normal, with mean λ and Standard deviation
If we standarize X, we get a random variable W
The values of W are tabulated and can be found on the attached file. We want a 95% confidence interval, so we want Z such that
P(-Z < w < Z) = 0.95
Using the symmetry of the normal density function, we get that
P(W < Z) = 0.975
If we look at the table, we will find that Z = 1.96, therefore we have
Equivalently,
Taking out the X and the sign, after reverting the inequalities, we obtain
Thus, a confidence interval with 95% confidence is
The (absolute) margin of error of this interval is we want that number to be at most 5, so we take n such that
We take n = 79. I hope that works for you!
