Questions: You are performing a study about weekly per capita milk consumption. A previous study found weekly per capita milk consumption to be normally distributed, with a mean of 41.9 fluid ounces a standard deviation of 7.6 fluid ounces. You randomly sample 30 people and record the weekly milk consumptions shown below. 33, 46, 31, 33, 39, 43, 29, 52, 36, 42, 37, 27, 36, 46, 38, 37, 37, 41, 44, 44, 36, 31, 25, 39, 52, 38, 43, 37, 45, 34 Find the mean of your sample. The mean is 38.4. (Round to one decimal place as needed.) Find the standard deviation of your sample. The standard deviation is 6.6. (Round to one decimal place as needed.) Compare the mean and standard deviation of your sample with those of the previous study. Discuss the differences.

Solution Steps

The sample mean \( \mu \) is calculated using the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

For the given data, we have:

\[ \mu = \frac{1151}{30} = 38.4 \]

Thus, the sample mean is \( 38.4 \).

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

For the sample, the variance is found to be \( 43.6 \). The standard deviation \( \sigma \) is then:

\[ \sigma = \sqrt{43.6} = 6.6 \]

Thus, the sample standard deviation is \( 6.6 \).

The previous study reported a mean of \( 41.9 \) fluid ounces and a standard deviation of \( 7.6 \) fluid ounces.

• The sample mean \( 38.4 \) is less than the previous mean \( 41.9 \), indicating that on average, consumption from the sample is less than in the previous study.
• The sample standard deviation \( 6.6 \) is less than the previous standard deviation \( 7.6 \) by:

\[ 7.6 - 6.6 = 1.0 \]

This indicates that the sample standard deviation is less spread out by \( 1.0 \) fluid ounces.

Final Answer