To find the mean and standard deviation of the weights, let's use the provided data points:

Weights: 29.3, 28.2, 29.4, 28.6, 28.6, 28.6 grams

Step 1: Calculate the Mean ($\bar{y}$)

The formula for the mean is: $\bar{y} = \frac{\sum y_i}{n}$ where $y_i$ are the individual weights and $n$ is the number of data points.

$\bar{y} = \frac{29.3 + 28.2 + 29.4 + 28.6 + 28.6 + 28.6}{6}$

Calculating the sum of weights: $29.3 + 28.2 + 29.4 + 28.6 + 28.6 + 28.6 = 172.7$

Now, divide by the number of weights (6): $\bar{y} = \frac{172.7}{6} = 28.78 \text{ grams}$

Step 2: Calculate the Standard Deviation (s)

The formula for the sample standard deviation is: $s = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n - 1}}$

1. Calculate each deviation from the mean, square it, and sum them up:

   • $(29.3 - 28.78)^2 = 0.2724$
   • $(28.2 - 28.78)^2 = 0.3364$
   • $(29.4 - 28.78)^2 = 0.3844$
   • $(28.6 - 28.78)^2 = 0.0324$
   • $(28.6 - 28.78)^2 = 0.0324$
   • $(28.6 - 28.78)^2 = 0.0324$

   Sum of squared deviations: $0.2724 + 0.3364 + 0.3844 + 0.0324 + 0.0324 + 0.0324 = 1.0904$

2. Divide by $n - 1 = 5$: $\frac{1.0904}{5} = 0.21808$

3. Take the square root: $s = \sqrt{0.21808} \approx 0.47 \text{ grams}$

Answers:

• Mean ($\bar{y}$) = 28.78 grams
• Standard deviation ($s$) = 0.47 grams

Would you like further details or explanations?

Here are some additional related questions you might find interesting:

1. How would the mean change if an additional bag with a weight of 29.0 grams was added?
2. What is the variance of these weights, and how does it relate to the standard deviation?
3. How would you interpret the standard deviation in the context of this data?
4. If the standard deviation was much higher, what could that indicate about the packaging process?
5. How does sample size affect the accuracy of the mean and standard deviation?

Tip: When working with small sample sizes, the sample standard deviation ($s$) provides a better estimate of population variability than using the population formula.