Question

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 ________.

Answer 1

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:

$X \sim N(208.19,53.44)$  

Where $\mu=208.19$ and $\sigma=53.44$

We select a sample size of n =35.

We know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And for this case the standard error is given by:

$\sigma_{\bar x}= \frac{53.44}{\sqrt{35}}= 9.033$

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

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:

$X \sim N(208.19,53.44)$  

Where $\mu=208.19$ and $\sigma=53.44$

We select a sample size of n =35.

We know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And for this case the standard error is given by:

$\sigma_{\bar x}= \frac{53.44}{\sqrt{35}}= 9.033$

Answer 2

Answer:

The standard error of the mean for this sample is 9.033

Step-by-step explanation:

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

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:

$\sigma = 53.44, n = 35$

So

$s = \frac{53.44}{\sqrt{35}} = 9.033$

The standard error of the mean for this sample is 9.033