Techy detail, how the range for the NET row is calculated in the section; "TECHY DETAILS - IF YOU ADD UP ALL THE WAYS HEAT CAN BE TRAPPED, ASSUME ALL ARE THE WORST THEY CAN BE, THEY AREN’T ENOUGH TO OVERCOME THE EXTRA HEAT THE WORLD RADIATES JUST BY BEING WARMER - SO THERE CAN’T BE A RUNAWAY"
This is consistent with their "virtually certain" conclusion.
Yes, if you add up the lowest values and the highest values you get: -2.57 to 0.29. But these are independent variables.
In the IPCC calibrate language, "very likely" means that it is 90% certain it is within that range or 10% chance it is outside for each single variable.
Because they are independent, to be outside that range for all 5 variables is 10% of 10% of 10% of 10% of 10%.
Assuming it is equally likely to be above or below, for them all to be above 0.29 is 5% of 5% of 5% of 5% of 5%
There is only one chance in 20^5 or only one chance in 3.2 million of them all being above 0.29 on the positive side.
The simplest way is to assume they are all "normal" distributions. Then you can calculate it like this:
For the range -3.4 to -3, the mean is -3.2 and the 90% range is 0.2.
For the range 1.1 to 1.5, the mean is 1.3 and the 90% range is 0.2.
For the range 0.1 to 0.6, the mean is 0.35 and the 90% range is 0.25.
For the range -0.1 to 0.94, the mean is 0.42 and the 90% range is 0.52.
For the range -0.27 to 0.25, the mean is -0.01 and the 90% range is 0.26.
Sum of the means is -3.2+1.3+0.35+0.42-0.01 = -1.14.
90% of values fall within 1.645 standard deviations
The standard deviation of the sum is sqrt (sum of squares of standard deviations)
Techy detail, how the range for the NET row is calculated in the section; "TECHY DETAILS - IF YOU ADD UP ALL THE WAYS HEAT CAN BE TRAPPED, ASSUME ALL ARE THE WORST THEY CAN BE, THEY AREN’T ENOUGH TO OVERCOME THE EXTRA HEAT THE WORLD RADIATES JUST BY BEING WARMER - SO THERE CAN’T BE A RUNAWAY"
This is consistent with their "virtually certain" conclusion.
Yes, if you add up the lowest values and the highest values you get: -2.57 to 0.29. But these are independent variables.
In the IPCC calibrate language, "very likely" means that it is 90% certain it is within that range or 10% chance it is outside for each single variable.
Because they are independent, to be outside that range for all 5 variables is 10% of 10% of 10% of 10% of 10%.
Assuming it is equally likely to be above or below, for them all to be above 0.29 is 5% of 5% of 5% of 5% of 5%
There is only one chance in 20^5 or only one chance in 3.2 million of them all being above 0.29 on the positive side.
The simplest way is to assume they are all "normal" distributions. Then you can calculate it like this:
For the range -3.4 to -3, the mean is -3.2 and the 90% range is 0.2.
For the range 1.1 to 1.5, the mean is 1.3 and the 90% range is 0.2.
For the range 0.1 to 0.6, the mean is 0.35 and the 90% range is 0.25.
For the range -0.1 to 0.94, the mean is 0.42 and the 90% range is 0.52.
For the range -0.27 to 0.25, the mean is -0.01 and the 90% range is 0.26.
Sum of the means is -3.2+1.3+0.35+0.42-0.01 = -1.14.
90% of values fall within 1.645 standard deviations
The standard deviation of the sum is sqrt (sum of squares of standard deviations)
= sqrt ( (0.2/1.645)^2 + (0.2/1.645)^2 + (0.25/1.645)^2 + (0.52/1.645)^2 + (0.26/1.645)^2)
= 0.42138667053
so the likely range of the sum is
-1.14 + - 1.645*0.42138667053
or -1.14 - 1.645*0.42138667053 to -1.14 + 1.645*0.42138667053
or -1.83 to -0.45
That is quite close to what they have there, but it's not quite the same.
Which I wasn't expecting anyway because they said the most likely value for the Plank is -3.22 while the mean is -3.2 so it's clearly a bit skewed.
So you don't expect it to be exactly the same but it's in the right ballpark
We can also work out the chance of it going above 0 from a likely range of -1.81 to -0.51
That is -1.16 +- 0.65
So the standard deviation is 0.65/1.645 = 0.395
So 0 is 1.16/0.395 = 2.94 standard deviations away from the mean.
Using a normal distribution calculator the probability of being within that is 0.99836
So the chance of being outside is (1-0.99836)/ 2 or 0.00082 or 0.082%
or 1 in 1/0.00082 or less than 1 in 1,200
That is assuming normal distribution when the distributions are clearly skewed or they wouldn't add up like that.
But it is much less than 1% anyway and probably 1 in 1000 or so.
This is all consistent with their "virtually certain" conclusion.