Basic statistics, calculating odds over time
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Given the data that in 2016 4.6 million people were seriously injured in car accidents I came up with the following over time assuming nothing else changes:
4.6M injured / 330M total population = 1.39% yearly risk of injury
1.39% risk of injury x 80 years avg lifespan = 111.5% risk of injury
But it is recommended to do it this way with the odds of not being injured:
(1 - 0.0139) ^ 80 = 32.6%
making the chance of serious injury 67.4% over 80 years.
Why does the first way not work?
statistics
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RobC
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Given the data that in 2016 4.6 million people were seriously injured in car accidents I came up with the following over time assuming nothing else changes:
4.6M injured / 330M total population = 1.39% yearly risk of injury
1.39% risk of injury x 80 years avg lifespan = 111.5% risk of injury
But it is recommended to do it this way with the odds of not being injured:
(1 - 0.0139) ^ 80 = 32.6%
making the chance of serious injury 67.4% over 80 years.
Why does the first way not work?
statistics
asked yesterday
RobC
204
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given the data that in 2016 4.6 million people were seriously injured in car accidents I came up with the following over time assuming nothing else changes:
4.6M injured / 330M total population = 1.39% yearly risk of injury
1.39% risk of injury x 80 years avg lifespan = 111.5% risk of injury
But it is recommended to do it this way with the odds of not being injured:
(1 - 0.0139) ^ 80 = 32.6%
making the chance of serious injury 67.4% over 80 years.
Why does the first way not work?
statistics
asked yesterday
RobC
204
Given the data that in 2016 4.6 million people were seriously injured in car accidents I came up with the following over time assuming nothing else changes:
4.6M injured / 330M total population = 1.39% yearly risk of injury
1.39% risk of injury x 80 years avg lifespan = 111.5% risk of injury
But it is recommended to do it this way with the odds of not being injured:
(1 - 0.0139) ^ 80 = 32.6%
making the chance of serious injury 67.4% over 80 years.
Why does the first way not work?
statistics
statistics
asked yesterday
RobC
204
asked yesterday
RobC
204
edited yesterday
asked yesterday
RobC
204
asked yesterday
RobC
204
asked yesterday
RobC
204
204
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1 Answer
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What you have calculated is the average number of injuries over 80 years. If every year you have a probability of 0.0139 of getting injured, the average number of injuries you will sustain in 80 years is $80 times 0.0139 = 1.112$. On the other hand, the probability of not getting injured at all in 80 years is the product of the probabilities of not getting injured in year 1, year 2, ..., year 80 (assuming that these events occur independently), which is $(1-0.0139)^{80}$.
I recommend reading up on the binomial distribution.
answered yesterday
Aditya Dua
3855
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you have calculated is the average number of injuries over 80 years. If every year you have a probability of 0.0139 of getting injured, the average number of injuries you will sustain in 80 years is $80 times 0.0139 = 1.112$. On the other hand, the probability of not getting injured at all in 80 years is the product of the probabilities of not getting injured in year 1, year 2, ..., year 80 (assuming that these events occur independently), which is $(1-0.0139)^{80}$.
I recommend reading up on the binomial distribution.
answered yesterday
Aditya Dua
3855
add a comment |
up vote
1
down vote
accepted
What you have calculated is the average number of injuries over 80 years. If every year you have a probability of 0.0139 of getting injured, the average number of injuries you will sustain in 80 years is $80 times 0.0139 = 1.112$. On the other hand, the probability of not getting injured at all in 80 years is the product of the probabilities of not getting injured in year 1, year 2, ..., year 80 (assuming that these events occur independently), which is $(1-0.0139)^{80}$.
I recommend reading up on the binomial distribution.
answered yesterday
Aditya Dua
3855
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you have calculated is the average number of injuries over 80 years. If every year you have a probability of 0.0139 of getting injured, the average number of injuries you will sustain in 80 years is $80 times 0.0139 = 1.112$. On the other hand, the probability of not getting injured at all in 80 years is the product of the probabilities of not getting injured in year 1, year 2, ..., year 80 (assuming that these events occur independently), which is $(1-0.0139)^{80}$.
I recommend reading up on the binomial distribution.
answered yesterday
Aditya Dua
3855
What you have calculated is the average number of injuries over 80 years. If every year you have a probability of 0.0139 of getting injured, the average number of injuries you will sustain in 80 years is $80 times 0.0139 = 1.112$. On the other hand, the probability of not getting injured at all in 80 years is the product of the probabilities of not getting injured in year 1, year 2, ..., year 80 (assuming that these events occur independently), which is $(1-0.0139)^{80}$.
I recommend reading up on the binomial distribution.
answered yesterday
Aditya Dua
3855
answered yesterday
Aditya Dua
3855
answered yesterday
Aditya Dua
3855
answered yesterday
Aditya Dua
3855
3855
add a comment |
add a comment |
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