Math Problem Statement
Use the Empirical Rule to determine the percentage of M&Ms with weights more than 0.9120.912 gram. using the data set 0.87
0.91
0.89
0.83
0.91
0.88
0.90
0.88
0.88
0.94
0.91
0.95
0.91
0.88
0.88
0.92
0.82
0.86
0.86
0.87
0.86
0.89
0.84
0.85
0.90
0.86
0.87
0.93
0.89
0.79
0.93
0.84
0.90
0.86
0.93
0.91
0.87
0.82
0.84
0.84
0.88
0.88
0.85
0.83
0.90
0.86
0.84
0.87
Solution
The mean weight of the M&Ms in the given dataset is approximately , and the standard deviation is about .
For a weight of , the z-score is approximately . According to the Empirical Rule, about 68% of the data lies within 1 standard deviation of the mean (both below and above).
Since the z-score is close to 1, we know that roughly of the data should lie above one standard deviation from the mean (i.e., above 0.912 grams).
Would you like further details on the Empirical Rule or have any questions?
Here are five related questions:
- How would the percentage change if the weight was 1 standard deviation higher?
- What is the percentage of M&Ms weighing less than 0.87 grams?
- How is the z-score calculated, and why is it important in the empirical rule?
- Can the Empirical Rule apply to non-normal data distributions?
- What percentage of data falls within two standard deviations of the mean?
Tip: The Empirical Rule applies well to normally distributed data. When data is skewed, other statistical methods may provide more accurate insights.
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Math Problem Analysis
Mathematical Concepts
Statistics
Empirical Rule
Z-scores
Normal Distribution
Formulas
Z-score formula: z = (x - μ) / σ
Empirical Rule percentages: 68%-95%-99.7% rule for 1, 2, and 3 standard deviations
Theorems
Empirical Rule (68%-95%-99.7%)
Normal Distribution properties
Suitable Grade Level
Grades 9-12
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