ClearBags is an online company that sells packaging materials for photographers. Suppose their average order size is $208.19 with a standard deviation of $53.44. A random sample of 35 customer orders has been selected. The standard error of the mean for this sample is ________.

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\sigma = 53.44, n = 35[/tex]

So

[tex]s = \frac{53.44}{\sqrt{35}} = 9.033[/tex]

The standard error of the mean for this sample is 9.033

dfbustos dfbustos · Answer 2 · 2020-03-29T03:28:12+02:00

Answer:

Let X the random variable that represent the order size of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(208.19,53.44)[/tex]

Where [tex]\mu=208.19[/tex] and [tex]\sigma=53.44[/tex]

We select a sample size of n =35.

We know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And for this case the standard error is given by:

[tex]\sigma_{\bar x}= \frac{53.44}{\sqrt{35}}= 9.033[/tex]

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem