mrs. juarez graded ten english papers and recorded the scores. 92, 95, 100, 62, 88, 90, 100, 96, 89, 98…

mrs. juarez graded ten english papers and recorded the scores. 92, 95, 100, 62, 88, 90, 100, 96, 89, 98 which statements are true? check all that apply. the range of scores is 38. without the outlier, the range of scores would be 12. the outlier impacts the range more than it impacts the interquartile range. the interquartile range is 9. the interquartile range is 4. without the outlier, the interquartile range would be 9.5.

mrs. juarez graded ten english papers and recorded the scores. 92, 95, 100, 62, 88, 90, 100, 96, 89, 98 which statements are true? check all that apply. the range of scores is 38. without the outlier, the range of scores would be 12. the outlier impacts the range more than it impacts the interquartile range. the interquartile range is 9. the interquartile range is 4. without the outlier, the interquartile range would be 9.5.

Answer

Explanation:

Step1: Arrange scores in ascending order

$62,88,89,90,92,95,96,98,100,100$

Step2: Calculate the range

Range = Maximum - Minimum. Maximum = 100, Minimum = 62. So, Range = $100 - 62=38$.

Step3: Identify the outlier

62 is an outlier as it is far - from the other values.

Step4: Calculate range without the outlier

New maximum = 100, new minimum = 88. New range = $100 - 88 = 12$.

Step5: Calculate quartiles

For $n = 10$ data points, the median is the average of the 5th and 6th ordered values. Median ($Q_2$)=$\frac{92 + 95}{2}=93.5$. The lower half is $62,88,89,90,92$. The median of the lower half ($Q_1$) = 89. The upper half is $95,96,98,100,100$. The median of the upper half ($Q_3$) = 98. Inter - quartile range (IQR)=$Q_3 - Q_1=98 - 89 = 9$.

Step6: Calculate IQR without the outlier

The new data set is $88,89,90,92,95,96,98,100,100$. For $n = 9$ data points, median ($Q_2$) = 95. The lower half is $88,89,90,92$. Median of lower half ($Q_1$)=$\frac{89 + 90}{2}=89.5$. The upper half is $96,98,100,100$. Median of upper half ($Q_3$)=$\frac{98+100}{2}=99$. New IQR=$99 - 89.5 = 9.5$.

Answer:

The range of scores is 38. Without the outlier, the range of scores would be 12. The outlier impacts the range more than it impacts the interquartile range. The interquartile range is 9. Without the outlier, the interquartile range would be 9.5.