Math Problem Statement

Students investigating the packaging of potato chips purchased 6 bags of chips marked with a net weight of 29.5 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.3, 28.2, 29.4, 28.6, 28.6, 28.6. Find the mean and standard deviation of the weights.

Solution

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 (yˉ\bar{y})

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

yˉ=29.3+28.2+29.4+28.6+28.6+28.66\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=172.729.3 + 28.2 + 29.4 + 28.6 + 28.6 + 28.6 = 172.7

Now, divide by the number of weights (6): yˉ=172.76=28.78 grams\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=(yiyˉ)2n1s = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n - 1}}

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

    • (29.328.78)2=0.2724(29.3 - 28.78)^2 = 0.2724
    • (28.228.78)2=0.3364(28.2 - 28.78)^2 = 0.3364
    • (29.428.78)2=0.3844(29.4 - 28.78)^2 = 0.3844
    • (28.628.78)2=0.0324(28.6 - 28.78)^2 = 0.0324
    • (28.628.78)2=0.0324(28.6 - 28.78)^2 = 0.0324
    • (28.628.78)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=1.09040.2724 + 0.3364 + 0.3844 + 0.0324 + 0.0324 + 0.0324 = 1.0904

  2. Divide by n1=5n - 1 = 5: 1.09045=0.21808\frac{1.0904}{5} = 0.21808

  3. Take the square root: s=0.218080.47 gramss = \sqrt{0.21808} \approx 0.47 \text{ grams}

Answers:

  • Mean (yˉ\bar{y}) = 28.78 grams
  • Standard deviation (ss) = 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 (ss) provides a better estimate of population variability than using the population formula.

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Math Problem Analysis

Mathematical Concepts

Statistics
Mean
Standard Deviation

Formulas

Mean formula: \( \bar{y} = \frac{\sum y_i}{n} \)
Sample Standard Deviation formula: \( s = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n - 1}} \)

Theorems

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Suitable Grade Level

Grades 10-12