General Betting
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23 Dec 10
09:46
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11 Oct 07
I came across this question by chance on a maths forum. I would be interested to see what the answer is generally thought to be. I have my view which doesn't match that given.
The journey to or from work has a mean of 67 minutes and a standard deviation of 15 minutes. You can assume a normal distribution of journey times.
a) What's the probability the workers total traveling time to and from work is more than 151 minutes?
The journey to or from work has a mean of 67 minutes and a standard deviation of 15 minutes. You can assume a normal distribution of journey times.
a) What's the probability the workers total traveling time to and from work is more than 151 minutes?
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0.211 assuming that the journey times are independent, which in a case like this is unlikely to be true.
Bayes is right. Mean for total journey is 134 mins, sd is 21.21. So prob = (in Excel) (1 - NORMDIST(151, 134, 21.21, TRUE))
The standard deviation comes from:
SD(X + X) = sqrt( Var(X) + Var(X))
The above only applies if the covariance of the two events (the trip to work and the trip home) is zero. If the times are independent then this is the case. Are the journey times of trips to and from work independent though? I doubt it!
SD(X + X) = sqrt( Var(X) + Var(X))
The above only applies if the covariance of the two events (the trip to work and the trip home) is zero. If the times are independent then this is the case. Are the journey times of trips to and from work independent though? I doubt it!
Thanks for your comments.
The experts answer is double the mean time and the SD to get the z score. So z=(151-134)/30 = 0.57, which gives .2843.
I do not think you can answer the question as worded, the journey times are not in, my view, independent.
Bayes, you have lost me (not difficult to do) with your SD calculation, and why, when you have a moment could you explain further.
The experts answer is double the mean time and the SD to get the z score. So z=(151-134)/30 = 0.57, which gives .2843.
I do not think you can answer the question as worded, the journey times are not in, my view, independent.
Bayes, you have lost me (not difficult to do) with your SD calculation, and why, when you have a moment could you explain further.
Equimine,
Your experts answer is wrong.
What Bayes is saying is basically that if you double the number of samples (in your example, 'samples' are time periods), then the SD increases by a factor of the square root of 2 (because SD is the root of variance).
Your experts answer is wrong.
What Bayes is saying is basically that if you double the number of samples (in your example, 'samples' are time periods), then the SD increases by a factor of the square root of 2 (because SD is the root of variance).
Contrarian & Bayes,
Thanks for that, I know the basic Bayes but wasn't aware of the point about the SD for multiple events.
As I said I don't think they independent events, but pleased to learn something new anyway.
Have a good Xmas.
Thanks for that, I know the basic Bayes but wasn't aware of the point about the SD for multiple events.
As I said I don't think they independent events, but pleased to learn something new anyway.
Have a good Xmas.
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