SOLUTION: A population of values has a normal distribution with mean = 173 and standard deviation = 55. You intend to draw a random sample size of n = 173 What is the mean of the distrib

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Question 1185733: A population of values has a normal distribution with mean = 173 and standard deviation = 55. You intend to draw a random sample size of n = 173
What is the mean of the distribution of sample means?

What is the standard deviation of the distribution of sample means? (Round 2 decimal places)
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the population mean is 173
the population standard deviation is 55
the sample size is 173

the standard error is equal to the standard deviation divided by the square root of the sample size = 55/sqrt(173) = 4.181572567.

round to two decimal places to get a standard error of 4.18.

the standard error is the standard deviation of the distribution of sample means.

the mean of each sample taken will be within a certain range depending on the confidence interval.

if you take enough samples, the mean of all the samples will approach the mean of the population.

the larger the sample size, the closer to the population mean will the sample mean be.

as an example, assume the confidence interval is .99.
that means that 99% of the samples will have a mean that is within the confidence interval.

your population mean is 173.
your population standard deviation is 55

with a sample size of 173, the standard error is 55 / sqrt(173) = 4.18157.

99% of your sample means will be between 162.229 and 183.771



with a sample size of 1000, the standard error is 55 / sqrt(1000) = 1.73925.

99% of your sample means will be between 168.52 and 177.48.



with a sample size of 10,000, the standard error is 55 / sqrt(10000) = .55

99% of your sample means will be between 171.583 and 174.417



the spread of the sample means get less the larger the size of each sample.

i think the simple answer will be that the distribution of sample means will have a mean of 173 and a standard error of 4.18

hopefully, that's what you're looking for.