SOLUTION: ⦁ The two-way table below gives the thousands of commuters in Massachusetts in 2015 by transportation method and one-way length of commute.
Less than 15 minutes 15 - 29 min
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Question 1171674: ⦁ The two-way table below gives the thousands of commuters in Massachusetts in 2015 by transportation method and one-way length of commute.
Less than 15 minutes 15 - 29 minutes 30 - 44 minutes 45 - 59 minutes 60 or more minutes Total
Private vehicle 636 908 590 257 256 2647
Public transportation 9 54 96 62 108 329
Other means 115 70 23 7 7 222
Total 760 1032 709 326 371 3198
⦁ Find the probability that either the commuter used public transportation and/or the commuter had a commute of 60 or more minutes. (2 marks)
⦁ Given that the commuter used public transportation, find the probability that the commuter had a commute of 60 or more minutes. (2 marks)
Found 2 solutions by Boreal, math_tutor2020:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Private vehicle 636 908 590 257 256 2647
Public transportation 9 54 96 62 108 329
Other means. 115 70 23 7 7 222
Total 760 1032 709 326 371 3198
Public transportation has 329 (end of the 2nd row)
60 or more minutes is 371 (fourth column total)
That is 600
subtract where they intersect (double counting) which is 108.
There were 492 of 3198 (15.4%)
Given they used public transportation (329), number that they had commute of 60 or more min was 108. That probability is 32.8%.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
| Less than 15 minutes | 15 - 29 minutes | 30 - 44 minutes | 45 - 59 minutes | 60 or more minutes | Total |
Private vehicle | 636 | 908 | 590 | 257 | 256 | 2647 |
Public transportation | 9 | 54 | 96 | 62 | 108 | 329 |
Other means | 115 | 70 | 23 | 7 | 7 | 222 |
Total | 760 | 1032 | 709 | 326 | 371 | 3198 |
Problem 1)
Find the probability that either the commuter used public transportation and/or the commuter had a commute of 60 or more minutes. (2 marks)
Add up all the values in the "public transportation" row and all of the values in the "60 or more minutes" column:
9+54+96+62+108+256+7 = 592
note: don't add in the totals as they already account for the values listed above.
As a shortcut we could add the totals for the row and column: 329+371 = 700
Then we subtract off 108 because it was counted twice: 700-108 = 592
Either way, there are 592 people who either took public transportation, their commute time was 60 minutes or more, or both events occurred.
Divide 592 over the grand total in the lower right corner and we get (592)/(3,198) = 0.185 approximately
Answer: Approximately 0.185
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Problem 2) Given that the commuter used public transportation, find the probability that the commuter had a commute of 60 or more minutes. (2 marks)
Focus solely on the "public transportation" row. There are 108 people whose commute time was 60 or more minutes. There are 329 people total who took public transportation.
The probability of a commuter having a comute time of 60 or more minutes, given they took public transportation, is 108/329 = 0.328
Answer: Approximately 0.328
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