Chapter 4: Problem 1
The following masses were recorded for 12 different U.S. quarters (all given in grams): $$\begin{array}{llll} 5.683 & 5.549 & 5.548 & 5.552 \\ 5.620 & 5.536 & 5.539 & 5.684 \\ 5.551 & 5.552 & 5.554 & 5.632 \end{array}$$ Report the mean, median, range, standard deviation and variance for this data.
Short Answer
Step by step solution
Calculate the Mean
Sort and Identify the Median
Determine the Range
Calculate the Standard Deviation and Variance
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mean Calculation
In the exercise, we are given the masses of 12 quarters, and the sum of these masses is 66.100 grams. By dividing this sum by 12, which is the total number of quarters, we arrive at the mean value:
\[ \text{Mean} = \frac{66.100}{12} = 5.5083 \]
This mean tells us that, on average, each quarter weighs approximately 5.5083 grams. The mean is useful for providing an overall picture of what each value would be if they were all the same.
Median Calculation
For the given data sorted as: 5.536, 5.539, 5.548, 5.549, 5.551, 5.552, 5.552, 5.554, 5.620, 5.632, 5.683, 5.684, we observe that there are 12 numbers.
When the dataset has an even number of observations, as in this case, the median is the average of the two middle numbers. Here, the 6th and 7th numbers are both 5.552.
Thus, the median is calculated as:
\[ \text{Median} = \frac{5.552 + 5.552}{2} = 5.552 \]
This median indicates that it divides the dataset into two equal halves, providing insight into the distribution of the data.
Range Determination
For the masses of the quarters, the maximum value is 5.684 grams, and the minimum value is 5.536 grams. By subtracting the minimum from the maximum, we find the range:
\[ \text{Range} = 5.684 - 5.536 = 0.148 \]
A smaller range indicates that the values are closely clustered together, while a larger range would suggest more variability. In this case, the range is quite small, implying that the masses of the quarters are similar to one another.
Standard Deviation and Variance Calculation
- Variance quantifies the average of the squared deviations from the mean. It highlights the extent of variability within the dataset.
- Standard deviation, being the square root of variance, provides a more intuitive measure of dispersion in the same units as the data itself.
Sum of squared differences from the mean is 0.101571. Dividing by the number of observations, the variance is:
\[ \text{Variance} = \frac{0.101571}{12} \approx 0.008464 \]
Taking the square root of the variance gives the standard deviation:
\[ \text{Standard Deviation} = \sqrt{0.008464} \approx 0.092 \]
These metrics show the consistency in the quarter masses, with a smaller standard deviation signifying that the data points are closely clustered around the mean.