To calculate the mean weight of the cereal boxes from the provided frequency distribution, we'll follow these steps:

Steps:

1. Identify the midpoint ($x$) of each weight class.
2. Multiply the midpoint by the corresponding frequency ($f$) for each class to get the weighted sum.
3. Sum up all the $f \cdot x$ values.
4. Divide the total $f \cdot x$ by the total number of boxes ($\sum f$).

Table Analysis:

From the given table:

• Weight intervals: $15.8 \leq x < 16.1$, $16.1 \leq x < 16.4$, $16.4 \leq x < 16.7$, $16.7 \leq x < 17$, $17 \leq x < 17.3$.
• Frequencies: 13, 20, 83, 14, 16.

Calculations:

1. Find the midpoint ($x$) for each interval: $\text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}$

   • For $15.8 \leq x < 16.1$: $x = \frac{15.8 + 16.1}{2} = 15.95$
   • For $16.1 \leq x < 16.4$: $x = \frac{16.1 + 16.4}{2} = 16.25$
   • For $16.4 \leq x < 16.7$: $x = \frac{16.4 + 16.7}{2} = 16.55$
   • For $16.7 \leq x < 17$: $x = \frac{16.7 + 17}{2} = 16.85$
   • For $17 \leq x < 17.3$: $x = \frac{17 + 17.3}{2} = 17.15$

2. Multiply midpoint ($x$) by frequency ($f$):

   • $15.95 \times 13 = 207.35$
   • $16.25 \times 20 = 325.00$
   • $16.55 \times 83 = 1374.65$
   • $16.85 \times 14 = 235.90$
   • $17.15 \times 16 = 274.40$

3. Total weighted sum ($f \cdot x$): $207.35 + 325.00 + 1374.65 + 235.90 + 274.40 = 2417.30$

4. Total frequency ($\sum f$): $13 + 20 + 83 + 14 + 16 = 146$

5. Calculate the mean: $\text{Mean} = \frac{\text{Total Weighted Sum}}{\text{Total Frequency}} = \frac{2417.30}{146} \approx 16.56 \, \text{ounces}$

Final Answer:

The mean weight of the boxes of cereal is approximately 16.56 ounces.

Do you want further clarification or details on any step?

Related Questions:

1. What is the formula for weighted mean, and why is it used in grouped data?
2. How does increasing the number of intervals affect the accuracy of the mean calculation?
3. Can we calculate the standard deviation from the given data? How?
4. How is this method different from calculating the mean in ungrouped data?
5. Why is it necessary to use midpoints instead of any other value in grouped data?

Tip: When working with frequency distributions, always ensure that the intervals are non-overlapping and collectively exhaustive for accurate calculations.