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Updated on: 13 Sep 2025, 06:26
Originally posted by Sajjad1994 on 19 Feb 2021, 07:30.
Last edited by Bunuel on 13 Sep 2025, 06:26, edited 3 times in total.
Last edited by Bunuel on 13 Sep 2025, 06:26, edited 3 times in total.
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Dropdown 1: 30
Dropdown 2: 29
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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71% (02:48) correct
29%
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Based on the given information, use the drop-down menus to most accurately complete the following statements.
For the week in question, the median number of miles run by the survey respondents could have been .
For the week in question, the minimum possible value for the average (mean) number of miles run by the survey respondents is
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Dropdown 1: 30
Dropdown 2: 29
rocky620
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23 Feb 2021, 04:58
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Miles ran by 20 runners (lowest to highest)
<10, <10, 10-20, 10-20, 20-30, 20-30, 20-30, 30-40, 30-40, 30-40, 30-40, 30-40, 40-50, 40-50, 40-50, 40-50, 40-50, 50+, 50+, 50+
A. For the week in question, the median number of miles run by the survey respondents could have been_____________.
The 10th and 11th runner would have ran between 30-40 miles. Since, the minimum distance ran will be 30 and the maximum would be less than 40 (say 39.999)
The median distance will be between 30 & 39.999 (30 would be included). Only Option B matches this criteria.
A. 28
B. 30
C. 40
D. 42
B. For the week in question, the minimum possible value for the average (mean) number of miles run by the survey respondents is______________.
For the minimum average miles ran we need to assume that each runner ran only the minimum distance in the particular distance bracket.
Total distance ran = # of runners*Miles Ran
2*.1 = .2 (considering that the runners ran .1 mile, if they ran <10 miles)
2*10 = 20
3*20 = 60
5*30 = 150
5*40 = 200
3*50 = 150
Total = 580.2
Average Miles ran = \(\frac{580.2}{20}\) = 29.01 ~ 29. Option C
A. 7.5
B. 17
C. 29
D. 31
Option B & Option C respectively
<10, <10, 10-20, 10-20, 20-30, 20-30, 20-30, 30-40, 30-40, 30-40, 30-40, 30-40, 40-50, 40-50, 40-50, 40-50, 40-50, 50+, 50+, 50+
A. For the week in question, the median number of miles run by the survey respondents could have been_____________.
The 10th and 11th runner would have ran between 30-40 miles. Since, the minimum distance ran will be 30 and the maximum would be less than 40 (say 39.999)
The median distance will be between 30 & 39.999 (30 would be included). Only Option B matches this criteria.
A. 28
B. 30
C. 40
D. 42
B. For the week in question, the minimum possible value for the average (mean) number of miles run by the survey respondents is______________.
For the minimum average miles ran we need to assume that each runner ran only the minimum distance in the particular distance bracket.
Total distance ran = # of runners*Miles Ran
2*.1 = .2 (considering that the runners ran .1 mile, if they ran <10 miles)
2*10 = 20
3*20 = 60
5*30 = 150
5*40 = 200
3*50 = 150
Total = 580.2
Average Miles ran = \(\frac{580.2}{20}\) = 29.01 ~ 29. Option C
A. 7.5
B. 17
C. 29
D. 31
Option B & Option C respectively
19 Aug 2025, 01:10
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Part 2.
For the minimum average miles ran we need to assume that each runner ran only the minimum distance in the particular distance bracket.
Total = 580 (approx.)
Avg. mile ran would be = 580/20 = 29.
OPTION C
For the minimum average miles ran we need to assume that each runner ran only the minimum distance in the particular distance bracket.
Total = 580 (approx.)
Avg. mile ran would be = 580/20 = 29.
OPTION C
