## Poll

10 votes (45.45%) | |||

9 votes (40.9%) | |||

5 votes (22.72%) | |||

2 votes (9.09%) | |||

7 votes (31.81%) | |||

3 votes (13.63%) | |||

4 votes (18.18%) | |||

3 votes (13.63%) | |||

9 votes (40.9%) | |||

6 votes (27.27%) |

**22 members have voted**

March 30th, 2020 at 7:25:50 AM
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Quote:ssho8839671 ?

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.

March 30th, 2020 at 7:31:30 AM
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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

March 30th, 2020 at 8:43:37 AM
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I think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.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.

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.

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.

March 30th, 2020 at 8:57:43 AM
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Quote:charliepatrickI think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.

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.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.

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.

March 30th, 2020 at 9:05:34 AM
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Quote:ssho88Quote:charliepatrickI think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.

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.

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.

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.

March 30th, 2020 at 9:40:47 AM
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Quote:ThatDonGuyQuote:ssho88Quote:charliepatrickI think ssho88 has the figure the wrong way up (so to speak) but agree with his idea.

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.

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.

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

March 30th, 2020 at 11:05:56 AM
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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.

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.

March 30th, 2020 at 7:59:16 PM
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Here is the next one.

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

March 30th, 2020 at 8:29:03 PM
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Hey 19