Respuesta :
Answer:
The correct answers are "After adding the 0 test score, the mean would be affected.", ad "Before the missed test, Eva’s median score was 91".
Step-by-step explanation:
We know that the mean and median scores would not be helpful in determining Eva's abilities since the 0 is not a representation of how much she knows. However, it will affect the average as adding a 0 to the numerator and a 1 to the denominator will lower the average.
Finally, we can tell the median is 91 before added as when we line up the scores in ascending order, the middle number is 91.
82, 90, 91, 96, 100
Answer:
Statements that are true based on the data:
-After adding the 0 test score, the mean would be affected.
-After adding the 0 test score, the median would be the most appropriate measure of center to describe the data.
-Before the missed test, Eva’s median score was 91.
-Before the missed test, Eva’s mean score was 91.8.
Step-by-step explanation:
-After adding the 0 test score, the mean would be the most appropriate measure of center to describe the data: NO, it won't because the data now has an extreme value that shifts the mean value significantly.
-After adding the 0 test score, the mean would be affected: YES, The mean value would be greatly impacted by adding the 0, (mean=76.5 with the 0; 91.8 without the 0),
-After adding the 0 test score, the median would be the most appropriate measure of center to describe the data: YES, because the median won't be affected as much as the mean by the extreme value.
-Before the missed test, Eva’s median score was 96: NO, the median score was 91 at that point.
-Before the missed test, Eva’s median score was 91: YES, the median score was 91, the median is defined as the value where 50% of sorted values are below and 50% are above
-Before the missed test, Eva’s mean score was 91.8: YES, the mean is defined as the sum of all values divided by the total number of values
The formula for calculating the median is:
-odd number of values: [tex]$ x_m = x_\frac{n+1}{2}[/tex]
-even number of values: [tex]$ x_m = \frac{x_\frac{n}{2}+x_\frac{n+2}{2}}{2}[/tex]
MEDIAN BEFORE ADDING THE 0: 82,90,91,96,100
[tex]$ x_m = x_\frac{5+1}{2}[/tex]=3 (third value)
MEDIAN AFTER ADDING THE 0: 0,82,90,91,96,100,
[tex]$ x_m = \frac{x_\frac{6}{2}+x_\frac{6+2}{2}}{2}[/tex]=average between the 3 and 4 values
[tex]$ x_m = \frac{90+91}{2}[/tex]=90.5
The formula for calculating the mean is: [tex]mean ={\frac{1}{n} \sum_{i=1}^n (x_i) $[/tex]
MEAN BEFORE ADDING THE 0
[tex]mean ={\frac{1}{5} \sum_{i=1}^5 (x_i) $[/tex]=[tex]\frac{459}{5}[/tex]=91.8
MEAN AFTER ADDING THE 0
[tex]mean ={\frac{1}{6} \sum_{i=1}^6 (x_i) $[/tex]=[tex]\frac{459}{6}[/tex]=76.5