A normal population has an expected value of 60 and a variance of 30. Use the central limit theorem to determine what the sample size should be such that the mean has a probability of 90%
to fall between the values 58 and 62.
One concept of the Central Limit Theorem is something called the "standard error of the mean" which represents the standard deviation of the sampling distribution of the mean. This is calculated by taking the standard deviation in a population and dividing it by the square root of the sample size. Since the sample size is what you are trying to find here, you will need to set up equations to solve for that value. Note that standard deviation is the positive square root of the variance. If we use a z-table for normal distributions, we find the value +/- 1.645 to represent 90% for this problem. Setting up the equations:
-1.645 = (58 - 60)/(√30/√n)
1.645 = (62 - 60)/(√30/√n)
Solving either equation for n will yield the sample size. You can round the sample size to the next highest whole number if you are asked to do so.
Hopefully someone else can help you with your other two posts.
To solve for the sample size, we'll set up the equations using the z-score of +/-1.645, representing the 90% probability interval between 58 and 62.
Using the equation (-1.645) = (58 - 60) / (√30/√n), we can solve for n:
-1.645 = -2 / (√30/√n)
First, solve for the square root term:
(√30/√n) = -2 / -1.645
Next, cross-multiply to isolate the n term:
√n = (√30 * -1.645) / -2
Simplifying the right-hand side of the equation:
√n = √30 * 1.645 / 2
Then, square both sides to eliminate the square root:
(n) = [ (√30 * 1.645 / 2) ]^2
Calculate the right-hand side of the equation:
n = [ (1.732 * 1.645 / 2) ]^2
n = [ 1.732 * 0.8225 ]^2
n = 1.424481025^2
n ≈ 2.028
Rounding up to the next whole number, the sample size should be approximately 3.