Question

The distribution of actual weights of 8-ounce wedges of cheddar cheese produced at a dairy is Normal with mean 8.1 ounces and standard deviation 0.2 ounces. A sample of 10 of these cheese wedges is selected. The distribution of the sample mean of the weights of cheese wedges is:
approximately Normal, mean 8.1, standard deviation 0.2.

approximately Normal, mean 8.1, standard deviation 0.020.

It is not possible to tell because the sample size is too small.

approximately Normal, mean 8.1, standard deviation 0.063.

Answer 1

Answer:

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

$\bar X \sim N(\mu=8.1, \frac{0.2}{\sqrt{10}}=0.063)$

approximately Normal, mean 8.1, standard deviation 0.063.

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable weights of 8-ounce wedges of cheddar cheese produced at a dairy. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =8.1,\sigma =0.2)$

We take a sample of n=10 . That represent the sample size.

What can we say about the shape of the distribution of the sample mean?

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

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

$\bar X \sim N(\mu=8.1, \frac{0.2}{\sqrt{10}}=0.063)$

approximately Normal, mean 8.1, standard deviation 0.063.