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.

15.

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. What is the standard deviation of the sampling distribution of the mean?
0.075 ounces
0.963 ounces
0.315 ounces
0.0633 ounces

16.

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 company decides instead to sample batches of 20 cheese wedges, and the sampling is repeated every time workers start a new shift at the dairy. How will the distribution of the sample means of the weights of cheese wedges change from the previous batches, which only contained 10 samples?
The distribution will still be Normal, but it will be more peaked around the sample mean and the standard deviation will be larger.
The distribution will still be Normal, but it will be more peaked around the sample mean and the standard deviation will be smaller.
The shape of the distribution may change completely based on the new data.
It is not possible to tell from the information provided.

Answer 1

approximately Normal, mean 8.1, standard deviation 0.063

15.

0.063 ounces

16.

The distribution will still be Normal, but it will be more peaked around the sample mean and the standard deviation will be smaller.

Explanation:

As the distribution of the actual weights of 8-ounce wedges of cheddar cheese are normally distributed with mean μ=8.1 and standard deviation σ=0.2, so, according to central limit theorem the distribution of the sample mean of the weights of cheese wedges will be approximately Normal with mean μxbar and standard deviation σxbar.

μxbar=μ=8.1

σxbar=σ/√n=0.2/√10=0.2/3.1623=0.063

So, the distribution of the sample mean of the weights of cheese wedges is approximately Normal with mean 8.1 and standard deviation 0.063.

15.

For sample of size 10 the standard deviation of the sampling distribution of the mean is

σxbar=σ/√n=0.2/√10=0.2/3.1623=0.063

16.

The standard deviation of the sampling distribution of the mean for size 20 is

σxbar=σ/√n=0.2/√20=0.045.

and mean of the sampling distribution of the mean for size 20 will be 8.1.

The sampling distribution of mean will be normal with mean 8.1 and standard deviation 0.045 because the distribution of the actual weights of 8-ounce wedges of cheddar cheese are normally distributed.

We can see that for sample size 20 the standard deviation of the sampling distribution of the mean is smaller than for sample size 10 i.e. 0.045<0.063.

Thus, the distribution will still be Normal, but it will be more peaked around the sample mean and the standard deviation will be smaller.