Poll

7 votes (53.84%)
5 votes (38.46%)
3 votes (23.07%)
2 votes (15.38%)
6 votes (46.15%)
1 vote (7.69%)
2 votes (15.38%)
2 votes (15.38%)
7 votes (53.84%)
5 votes (38.46%)

13 members have voted

Wizard
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Wizard
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March 30th, 2020 at 7:25:50 AM permalink
Quote: ssho88

39671 ?



Shouldn't the mean needed be 20,000,000 / 500 = 40,000? Wouldn't they have a greater than 50% chance of running out with less? We're looking to have a 5% chance only of running out.
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
ssho88
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March 30th, 2020 at 7:31:30 AM permalink


mean = 1/500 = 0.002, Var = 0.002(assumed), SD =0.04472


For > 95% of chance, Z > -1.645
(X -40000)/0.04472/(20,000,000)^0.5 > -1.645
X > 39671

I am not sure about the Variance value . . .


Last edited by: ssho88 on Mar 30, 2020
charliepatrick
charliepatrick
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March 30th, 2020 at 8:43:37 AM permalink
I think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.
N=20m, p=1/500, q=499/500.
Average = Np = 40k.
SD = SQRT(N p q) = SQRT(20m / 500 * 499/500 ) = 199.7999. (Since q is nearly 1, this happens to be SQRT (Average) )
Number of SDs for 90% (i.e. 5% at either end) = 1.644854 ( https://en.wikipedia.org/wiki/Standard_deviation ).
So number (over average) needed is 328.642.

Hence number to ensure 95% = 40k + 328.642 = 40329.
If the government is going to buy the normal amount it might as well buy a few extra. The main problem is guessing whether 1 in 500 is correct in the first place. For instance if it was 1 in 496 then you'd now only have a 50% chance of having enough.
ssho88
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March 30th, 2020 at 8:57:43 AM permalink
Quote: charliepatrick

I think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.

N=20m, p=1/500, q=499/500.
Average = Np = 40k.
SD = SQRT(N p q) = SQRT(20m / 500 * 499/500 ) = 199.7999. (Since q is nearly 1, this happens to be SQRT (Average) )
Number of SDs for 90% (i.e. 5% at either end) = 1.644854 ( https://en.wikipedia.org/wiki/Standard_deviation ).
So number (over average) needed is 328.642.

Hence number to ensure 95% = 40k + 328.642 = 40329.
If the government is going to buy the normal amount it might as well buy a few extra. The main problem is guessing whether 1 in 500 is correct in the first place. For instance if it was 1 in 496 then you'd now only have a 50% chance of having enough.




I think we have the same mean and variance, mean and variance calculated by me is 40000 and 200 [0.04472 * (20,000,000)^0.5]

(X - 40000)/200 >-1.645

X > 39671

The difference is I used Z > -1.645 instead of +1.645 ??? Which one is correct ? I am not sure.
ThatDonGuy
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March 30th, 2020 at 9:05:34 AM permalink
Quote: ssho88

Quote: charliepatrick

I think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.

N=20m, p=1/500, q=499/500.
Average = Np = 40k.
SD = SQRT(N p q) = SQRT(20m / 500 * 499/500 ) = 199.7999. (Since q is nearly 1, this happens to be SQRT (Average) )
Number of SDs for 90% (i.e. 5% at either end) = 1.644854 ( https://en.wikipedia.org/wiki/Standard_deviation ).
So number (over average) needed is 328.642.

Hence number to ensure 95% = 40k + 328.642 = 40329.
If the government is going to buy the normal amount it might as well buy a few extra. The main problem is guessing whether 1 in 500 is correct in the first place. For instance if it was 1 in 496 then you'd now only have a 50% chance of having enough.




I think we have the same mean and variance, mean and variance calculated by me is 40000 and 200 [0.04472 * (20,000,000)^0.5]

(X - 40000)/200 >-1.645

X > 39671

The difference is I used Z > -1.645 instead of +1.645 ??? Which one is correct ? I am not sure.


Presumably, to be 95% confident, you find the value such that 90% of the values are within that many SDs on either side of the mean; 5% will be greater than the mean + that many SDs, so 95% are less than that value.
ssho88
ssho88
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March 30th, 2020 at 9:40:47 AM permalink
Quote: ThatDonGuy

Quote: ssho88

Quote: charliepatrick

I think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.

N=20m, p=1/500, q=499/500.
Average = Np = 40k.
SD = SQRT(N p q) = SQRT(20m / 500 * 499/500 ) = 199.7999. (Since q is nearly 1, this happens to be SQRT (Average) )
Number of SDs for 90% (i.e. 5% at either end) = 1.644854 ( https://en.wikipedia.org/wiki/Standard_deviation ).
So number (over average) needed is 328.642.

Hence number to ensure 95% = 40k + 328.642 = 40329.
If the government is going to buy the normal amount it might as well buy a few extra. The main problem is guessing whether 1 in 500 is correct in the first place. For instance if it was 1 in 496 then you'd now only have a 50% chance of having enough.




I think we have the same mean and variance, mean and variance calculated by me is 40000 and 200 [0.04472 * (20,000,000)^0.5]

(X - 40000)/200 >-1.645

X > 39671

The difference is I used Z > -1.645 instead of +1.645 ??? Which one is correct ? I am not sure.


Presumably, to be 95% confident, you find the value such that 90% of the values are within that many SDs on either side of the mean; 5% will be greater than the mean + that many SDs, so 95% are less than that value.






I agree. "to have a 95% chance of having enough" means LESS THAN that Z value. I interpreted it as MORE THAN in previous calculations.
Last edited by: ssho88 on Mar 30, 2020
Wizard
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Wizard
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March 30th, 2020 at 11:05:56 AM permalink
I agree with Charlie's answer.

The problem meant to ask how many ventilators should they have so that the probability of running out would be 5%, which is equivalent to a 95% chance of not running out.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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March 30th, 2020 at 7:59:16 PM permalink
Here is the next one.

It's not whether you win or lose; it's whether or not you had a good bet.
ChumpChange
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March 30th, 2020 at 8:29:03 PM permalink
Hey 19
Wizard
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March 30th, 2020 at 8:31:27 PM permalink
Quote: ChumpChange

Hey 19



I disagree
It's not whether you win or lose; it's whether or not you had a good bet.

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