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05 Sep 2024, 03:14
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The repair required less than half the average time to complete: True for all chips with spidering
The chip diameter was larger than the median chip diameter.: False for at least one chip with spidering
The insurance reimbursement was greater than or equal to the median insurance reimbursement.: False for at least one chip with spidering
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:43) correct
24%
(03:05)
wrong
based on 543
sessions
History
Date
Time
Result
Not Attempted Yet
On a recent workday, a technician repaired one chip—a small area from which a fragment of glass had been broken—in the windshield of each of 9 automobiles. For each automobile, the table indicates the chip’s diameter, in millimeters (mm); whether the chip was accompanied by spidering, which is a pattern of cracks radiating out from the chip; the volume, in milliliters (mL), of sealant used to repair the chip and any accompanying spidering; the time, in hours (h), to complete the repair; and the insurance reimbursement, which is the amount, in US dollars (US$), that was reimbursed by insurance.
For each of the following, select True for all chips with spidering if the given information indicates that the statement is true for each of the 9 chips that was accompanied by spidering. Otherwise, select False for at least one chip with spidering.
| Automobile | Chip diameter (mm) | Spidering? (yes/no) | Volume of sealant used (mL) | Time to complete repair (h) | Insurance reimbursement (US$) |
|---|---|---|---|---|---|
| 1 | 17 | yes | 5 | 0.25 | 45 |
| 2 | 8 | no | 2 | 0.75 | 52 |
| 3 | 24 | yes | 5 | 0.25 | 70 |
| 4 | 5 | no | 1 | 1 | 70 |
| 5 | 17 | yes | 4 | 0.25 | 52 |
| 6 | 17 | no | 3 | 0.75 | 45 |
| 7 | 12 | no | 2 | 1 | 70 |
| 8 | 10 | no | 2 | 0.5 | 70 |
| 9 | 6 | yes | 6 | 0.25 | 52 |
| Average (arithmetic mean) | 12.89 | - | 3.33 | 0.56 | 58.44 |
For each of the following, select True for all chips with spidering if the given information indicates that the statement is true for each of the 9 chips that was accompanied by spidering. Otherwise, select False for at least one chip with spidering.
| True for all chips with spidering | False for at least one chip with spidering | |
| The repair required less than half the average time to complete | ||
| The chip diameter was larger than the median chip diameter. | ||
| The insurance reimbursement was greater than or equal to the median insurance reimbursement. |
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Official Answer
The repair required less than half the average time to complete: True for all chips with spidering
The chip diameter was larger than the median chip diameter.: False for at least one chip with spidering
The insurance reimbursement was greater than or equal to the median insurance reimbursement.: False for at least one chip with spidering
08 Sep 2024, 01:23
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The best way to solve these questions is just have a glance of what is given
Variables and any combined aggregated values
- Variables - Diameter, Volume, Spidering (yes/no), Insurance reibursement
- Aggregated Values - Average
Lets go on option
Who Q is based on spidering - so lets quickly sort the table and any where there is mentioned NO spidering just move on.
Now,
We have 4 spidering options - lazer focus there.
A. Mean is already given in table so we dont need to calculate - great news. For all 4 spidering the time is less than half of mean (0.56/2). Click yes for all
B. Calculate median of all values in column diameter - 5,6,8,10,12,17,17,17,24 - Median is 12. One spidering one doesnt has it greater than 12 but rest do so choose Atleast one column
C. Calculate median of all values in column insurance - 45,45,52,52,52,70,70,70,70 - Median is 52. One spidering one doesnt has it greater than 52 but rest do so choose Atleast one column
Variables and any combined aggregated values
- Variables - Diameter, Volume, Spidering (yes/no), Insurance reibursement
- Aggregated Values - Average
Lets go on option
Who Q is based on spidering - so lets quickly sort the table and any where there is mentioned NO spidering just move on.
Now,
We have 4 spidering options - lazer focus there.
A. Mean is already given in table so we dont need to calculate - great news. For all 4 spidering the time is less than half of mean (0.56/2). Click yes for all
B. Calculate median of all values in column diameter - 5,6,8,10,12,17,17,17,24 - Median is 12. One spidering one doesnt has it greater than 12 but rest do so choose Atleast one column
C. Calculate median of all values in column insurance - 45,45,52,52,52,70,70,70,70 - Median is 52. One spidering one doesnt has it greater than 52 but rest do so choose Atleast one column
23 Sep 2024, 03:37
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im not able to understand the question can anbody explain this
