The distribution of student loan amounts for students is normally distributed ranging from zero to $100,000. Use your knowledge of the Empirical Rule to approximate the standard deviation of the dataset?
A) approximately $21,000
B) approximately $12,500
C) approximately $19,700
D) approximately $16,600
SD = approx. range/4
To approximate the standard deviation of the dataset using the Empirical Rule, we need to remember that the Empirical Rule states that:
- In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Since the dataset in this question ranges from zero to $100,000, we can assume that it is a continuous distribution and that the mean is at the middle of this range, i.e., $50,000.
To find the standard deviation, we can use the empirical rule as follows:
- For one standard deviation, approximately 68% of the data falls within this range. Since the range is $100,000, we can estimate it as $100,000 - $0 = $100,000. Therefore, one standard deviation (approximately 68% range) will be $100,000 / 6.
- For two standard deviations, approximately 95% of the data falls within this range. So, we can estimate it as two times the range for one standard deviation, i.e., $100,000 / 3.
- For three standard deviations, approximately 99.7% of the data falls within this range. So, we can estimate it as three times the range for one standard deviation, i.e., $100,000 / 2.
Now, we can compare these ranges with the answer choices:
A) approximately $21,000: This is not within the range estimated for one standard deviation.
B) approximately $12,500: This is within the range estimated for one standard deviation.
C) approximately $19,700: This is not within the range estimated for one standard deviation.
D) approximately $16,600: This is within the range estimated for one standard deviation.
Therefore, the most accurate answer choice for the standard deviation of the dataset would be B) approximately $12,500.