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12 May 2025, 01:36
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Dropdown 1: 0.05
Dropdown 2: 0.35
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:
77% (02:23) correct
23%
(02:42)
wrong
based on 323
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Not Attempted Yet
For a small city, the graph represents the daily deviation, in minutes, of a commuter’s travel time from the expected commute time for each day in a 100-day period. Data is grouped into disjoint classes of deviations: for each value of T marked on the horizontal axis, the class centered at T includes all observed deviations greater than or equal to (T−1) minutes but less than (T+1) minutes. The height of each bar represents the number of deviations in the corresponding class. A given day’s commute is said to be x minutes shorter than normal if it is x minutes less than the left endpoint of the class centered at 0, and x minutes longer than normal if it is x minutes greater than the right endpoint of the class centered at 0.
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
For a randomly selected day in this 100-day period, the probability that the commute time was at least 5 minutes longer than normal is , and the probability that the commute time was at least 1 minute longer than normal is .
GMAT-Club-Forum-6dgokq2c.png [ 23.44 KiB | Viewed 4127 times ]
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
For a randomly selected day in this 100-day period, the probability that the commute time was at least 5 minutes longer than normal is , and the probability that the commute time was at least 1 minute longer than normal is .
Attachment:
GMAT-Club-Forum-6dgokq2c.png [ 23.44 KiB | Viewed 4127 times ]
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Official Answer
Dropdown 1: 0.05
Dropdown 2: 0.35
14 May 2025, 00:44
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Shouldn't the question say 100 day period instead of 90? really threw me off for a second there.
20 May 2025, 01:30
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Bunuel
Solution:
Class centered at 0 includes deviations from -1 to +1 minutes.
1. Probability that commute time was at least 5 minutes longer than normal:
Corresponds to:
Class centered at 6 → 5 days
Total = 17 days
Probability = 5 / 100 = 5%
2. Probability that commute time was at least 1 minutes longer than normal:
Corresponds to:
Class centered at 2 → 18 days
Class centered at 4 → 12 days
Class centered at 6 → 5 days
Total = 18 + 12 + 5 = 35 days
Probability = 35 / 100 = 35%
