Answer :
Final answer:
The flipped course has more dispersion in exam scores using the range, while the traditional course has more dispersion using the sample standard deviation. Incorrectly recording a score significantly affects the range calculation.
Explanation:
Range is the measure of dispersion calculated as the highest score minus the lowest score in a dataset. The traditional course has a range of 36.4, while the flipped course has a range of 43.4. Therefore, the flipped course has more dispersion in exam scores using the range as the measure of dispersion.
Sample standard deviation measures the dispersion of scores around the mean. The traditional course has a standard deviation of 10.609, and the flipped course has a standard deviation of 8.128. Hence, the traditional course has more dispersion in exam scores using the sample standard deviation as the measure of dispersion.
If the score of 59.1 in the traditional course was incorrectly recorded as 591, the range calculation would change significantly.
The initial range was 36.4, but with the mistaken score, the new range would be 533.4, indicating a substantial impact on the dispersion of scores.
A) The traditional course has more dispersion in exam scores using the range as the measure of dispersion. B) The traditional course has more dispersion in exam scores. C) The range would change dramatically from [tex]\( 36.6 \) to \( 535 \)[/tex] due to the large discrepancy caused by the recording error.
Let's address each part of the question step by step.
Part (a): Range as Measure of Dispersion
First, we need to calculate the range for each course.
Traditional Course:
The exam scores are:
70.3, 68.5, 79.9, 67.6, 85.4, 79.1, 56.0, 80.4, 79.8, 71.7, 63.1, 70.4, 59.1, 75.5, 71.7, 63.5, 71.4, 77.7, 92.4, 79.6, 77.5, 83.0, 69.2, 92.6, 78.0, 76.9
Range for traditional course:
[tex]\[ \text{Range} = \max(\text{scores}) - \min(\text{scores}) = 92.6 - 56.0 = 36.6 \][/tex]
Flipped Course:
The exam scores are not provided directly, but the range is given as 27.2.
Answer for Part (a):
The traditional course has a range of 36.6 , while the flipped course has a range of 27.2 . Therefore, the traditional course has more dispersion in exam scores using the range as the measure of dispersion.
Part (b): Sample Standard Deviation as Measure of Dispersion
Next, we need to calculate the sample standard deviation for each course.
Traditional Course:
Using the provided scores:
[tex]\[ \bar{x} = \frac{1}{25} \sum_{i=1}^{25} x_i = \frac{1844.6}{25} = 73.784 \][/tex]
Calculate the sum of squares of deviations:
[tex]\[ \sum_{i=1}^{25} (x_i - \bar{x})^2 = 5749.6784 \][/tex]
Sample variance:
[tex]\[ s^2 = \frac{5749.6784}{24} = 239.5691 \][/tex]
Sample standard deviation:
[tex]\[ s = \sqrt{239.5691} = 15.478 \][/tex]
Flipped Course:
The standard deviation is given as 11.905.
Answer for Part (b):
The traditional course has a standard deviation of 15.478 , while the flipped course has a standard deviation of 11.905 . Therefore, the traditional course has more dispersion in exam scores using the sample standard deviation as the measure of dispersion.
Part (c): Effect of Recording Error
If the score of 59.1 in the traditional course was incorrectly recorded as 591, this error would significantly affect the range calculation.
Original scores:
70.3, 68.5, 79.9, 67.6, 85.4, 79.1, 56.0, 80.4, 79.8, 71.7, 63.1, 70.4, 59.1, 75.5, 71.7, 63.5, 71.4, 77.7, 92.4, 79.6, 77.5, 83.0, 69.2, 92.6, 78.0, 76.9
If 59.1 is recorded as 591:
{New scores} = 70.3, 68.5, 79.9, 67.6, 85.4, 79.1, 56.0, 80.4, 79.8, 71.7, 63.1, 70.4, 591, 75.5, 71.7, 63.5, 71.4, 77.7, 92.4, 79.6, 77.5, 83.0, 69.2, 92.6, 78.0, 76.9
New range:
[tex]\[ \text{New range} = \max(\text{scores}) - \min(\text{scores}) = 591 - 56.0 = 535 \][/tex]
Answer for Part (c):
The range would change dramatically from [tex]\( 36.6 \) to \( 535 \)[/tex] due to the large discrepancy caused by the recording error.
