Better safe than sorry?
posted at 2:05 pm on August 30, 2011 by Ed Morrissey
Allahpundit gave his thoughts on the backlash to Hurricane Irene coverage last night, from the perspective of someone who was in the path of the storm. I spent the hurricane over a thousand miles away from any of the damage, but I agree entirely with AP’s conclusion that while Irene may not have done the damage predicted before it hit the East Coast, the damage it did do justified the attention and the warnings issued. In my column for The Week, written before I read AP’s take, I argue that overprepared beats underprepared:
There is a legitimate concern about the danger of hyperbole in crisis-management situations. If every storm that approaches an American shore gets billed as another Katrina — a comparison often heard over the past week — the people who warnings are intended to help will start disregarding them. We saw the reluctance of residents in Irene’s path to follow evacuation requests; if they hear dire warnings of disasters that fail to materialize, then compliance with safety measures and evacuation orders will decrease, and put people at more risk. Hyperbolic estimates of damage and lack of essential services can also prompt unnecessary hoarding and artificial shortages of water and food that will end up making those goods both more expensive and less available even after an event-free storm.
Fortunately, the damage done by Irene came is far less than predicted — but the damage is not insignificant. Initial estimates of economic value lost have come to $7 billion, and that may go up as flooding continues in some areas. More significantly, at least 38 people died in the storm — as far south as Florida and as far west as Pennsylvania. The victims include an 89-year-old Connecticut woman who died when downed power lines set her home on fire, a New Jersey EMT who died in a Princeton flood, and a middle-aged New York man who had tried to rescue a child in a flood and got electrocuted by power lines.
Clearly, this was not a “manufactured” event. Irene may not have packed the punch that many predicted, but for those families and communities across 11 states who have to bury their dead and repair their homes, it wasn’t merely a photo opportunity. It was a real disaster, even if its scope was much more limited than initially feared.
Furthermore, we have a “dog that didn’t bark” dimension to this story. The storm was bad enough to kill dozens of people across 11 states. Without the warnings and the hyperbole, would the death toll have gone higher? None of the deaths appear to have resulted from excess zeal to seek safety or shelter. In fact, a number of them came from people who continued their recreational activities despite the storm. How many more were convinced to stay home instead?
Frankly, I’m a bit mystified about the complaints over saturation coverage. It’s not as if there weren’t other options, even for news junkies. As the storm approached, the coverage I saw reflected the loss of strength in the hurricane. It didn’t require as many breathless updates as the cable news networks provided, perhaps, but that’s a criticism that applies to cablers in general.
Much of the criticism went to the public demonstrations of leadership by executives like NYC Mayor Mike Bloomberg, New Jersey Governor Chris Christie, and Barack Obama. After Katrina, does anyone expect politicians to play it casual when hurricanes approach American shores? Besides, as I write in my column, even if one believes that these three were grandstanding at times — for which a fair argument can be made — it’s not going to make much difference for any of them politically, even in the short run. Americans expect executives to publicly demonstrate leadership, which means that all they get is a pass from the voters, not a bump, when things go right. Only in extreme circumstances, like Rudy Giuliani in New York City on 9/11, do politicians get long-term benefit from public demonstration of crisis management. Otherwise, these situations are mostly about not doing damage, both in real terms and to political reputations.
This storm did do real damage, and killed dozens of people. We should just be happy that it didn’t turn out as bad as we’d feared.
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Hurricanes form off Africa’s east coast? Uh, no.
Ronnie on August 31, 2011 at 4:57 AM
While the specific deaths due to traffic accidents can be predicted, the occurrence of traffic accidents, and the number of deaths attributable to such accidents, can most certainly be predicted statistically using a baseline occurrence of traffic accidents in the previous period. Are you really this stupid?
In fact, it’s much more difficult to predict deaths due to hurricane accidents. And certainly you agree that deaths due to hurricanes are accidents, right? You’re not claiming that people purposefully die during a hurricane, right? You do realize that such deaths are considered accidental deaths, right?
blink on August 31, 2011 at 8:27 AM
Wrong, right, wrong. Wrong,I did not skip a logical step. Right, NHC did not claim to be EXPECTING disaster. Wrong regarding REASONABLE possibility: NHC provided maps with zones of surge probabilities – separate maps showing probability zones for storm surges of 1,2,3,4,5,6 feet. I saw them. There were areas from the Hudson, around the battery, up the East River, thru Hell Gate, past City Island, and beyond where 5 foot storm surges had a probability of 10% or more. “Reasonable” is a subjective and contextual term, and I consider 10% to be a reasonable possibility in this case.
I’ll take your word for it.
I think we would agree that anyone who used the expression “better safe than sorry” to justify any level of over-preparedness, even if such level was irrational, would be thinking irrationally. Most people are usually rational, in my experience.
No, but you have demanded data in various other comments on this thread.
Yes I understand the difference. That’s why I used both terms; I was not trying to be redundant; that’s why I wrote “demands that everything be quantified and proven with data”. Sorry if I did not make the distinction clear.
In a normal distribution, the expected value is the point at which the integral represents 50% of the total area under the curve. Google it… I did, just to be sure I wasn’t talking off the side of my head like a complete idiot.
No, a confidence interval is something else entirely. Google it.
What’s sad is you telling me to Google it when you did not do so yourself, or were not knowledgeable enough to understand what you read.
What it means is that the area underneath the probablity curve of getting five or fewer 6′s is approximately 50%. I say approximately because you have chosen a discrete, non-continuous function.
Quit pretending that you know what you are talking about.
Look up the definition of “mode.” It is defined as the value that occurs (or is expected to occur) most frequently in a data set or a probability distribution and is considered the “likely” outcome – even if the probability of such outcome is significantly less than 50%.
The mode is the most frequent value, but not the expected value. Mode is a measure of actual data, not probability. If we have a data set of 1,1,1,1,6. The mode is 1, the mean, which estimates the expected value, is 2. If we are rolling dice, the expected value is 3.5, which is neither likely nor the mode, since it is impossible to roll a 3.5. But it is the expected value. Google it.
Yes it does. Likely means having a high probability. BY DEFINITION. GOOGLE IT!!!!!!!
It is the most likely because it has the highest probability. BY DEFINITION. GOOGLE IT!!!!!!!
“Stochastic” does not refer to the number of possible outcomes. Flipping a coin is a stochastic process, even though there are only two possible outcomes. BY DEFINITION. GOOGLE IT!!!!!!!
On this, we agree 100%. But one must also factor in averseness to risk. Otherwise, there would be no reason to purchase insurance, for reasons you describe.
Answer: No, I would not. I can afford to absorb a $100 lost. I would pay $1,000 to insure against a 1% chance of losing $90,000. In this case, better safe than sorry.
You clearly have zero understanding of how I am looking at this concept.
topdog on August 31, 2011 at 12:05 PM
Sorry, error in block quote formatting.
Here is a correction of middle section:
No. It doesn’t.
The mode is the most frequent value, but not the expected value. Mode is a measure of actual data, not probability. If we have a data set of 1,1,1,1,6. The mode is 1, the mean, which estimates the expected value, is 2. If we are rolling dice, the expected value is 3.5, which is neither likely nor the mode, since it is impossible to roll a 3.5. But it is the expected value. Google it.
Yes it does. Likely means having a high probability. BY DEFINITION. GOOGLE IT!!!!!!!
It is the most likely because it has the highest probability. BY DEFINITION. GOOGLE IT!!!!!!!
topdog on August 31, 2011 at 12:22 PM
I’m not sure what else needs to be said. After admitting this, then how can you claim that the disaster level reporting was justified?
Media coverage went completely beyond this.
I think I asked for data once in response to someone asking me for data. Regardless, it’s strange that you would accuse me of this in response to a question which didn’t ask you for data.
Are you claiming that damage level predictions from the hurricane were a normal distribution? I’ve repeated told you that the distribution is skewed – that should have made you realize that it wasn’t normal.
No, a 50% confidence interval is exactly what you described.
Ha ha ha ha ha!
1. “[F]ive or fewer” is NOT five! The expected number of times a 6 would be rolled is five. The probability of rolling five 6′s is NOT 50%. Just admit that your definition was pathetically inaccurate.
2. Your statement regarding “five or fewer” being 50% is inaccurate as well.
Only an idiot wouldn’t “expect” the most frequent outcome.
When building a model, such as a hurricane model, actual data is used. Probabilities built into these models are based on the actual data. The mode is relevant. Expectation is relevant.
Again, no it doesn’t. The “likely” outcome is the outcome which would be most expected.
EXACTLY!!! AND THE HIGHEST PROBABILITY IS LESS THAN 50%. This should tell you that likely doesn’t need to exceed 50%.
That should have read, “…a stochastic system when there are…”
1. Aversion to risk doesn’t need to be considered in prudent risk management. Aversion to risk is a reason to modify what prudent risk management suggests.
2. There are plenty of reasons for purchasing insurance even for risk tolerant decision makers.
EXACTLY! So quantifying the downside risk must be a factor in determining the level of resources expended to mitigate the risk.
Really? Would you pay $25,000 to insure against a 1% chance of losing $90,000? Remember, better safe than sorry!
blink on August 31, 2011 at 2:08 PM
If they went beyond reporting POSSIBILITY of disaster, then I agree with you. If that is what you experienced, I suggest changing networks. My experience was that they were forecasting a danger, not a certainty.
No, I agree the distribution is not “normal”. However, a normal distribution is frequently used as an approximation, just as economic theory assumes perfect markets: for simplification. I will be happy to split hairs for you. My point remains: in statistical language, the expected value is not necessarily the most likely value. Expected has a very specific meaning.
Perhaps you would like to go on Wikipedia and dispute with them the actual meaning of “expected” in statistics.
Note that this value is not necessarily the most likely value, and in the example of rolling dice, it is not even a possible value. Rolling a 3.5 would certainly be unexpected, but statistically speaking it is the expected value.
A 50% confidence interval ALSO relates to the integral of the probability curve, but with DIFFERENT limits. A coinfidence interval has limits that are the lower and upper boundaries of the confidence interval. The expected value integral has a lower limit of negative infinity (or zero, as the case may be) and an upper limit of the expected value.
Is that insane laughter?
Clear grasp of the obvious, at least sometimes.
My definition coincides with Wikipedia. Can you do better?
As I said, it is an approximation, and it is not precise because this is not a continuous function. I don’t care to do the math, but I would guess that the probability of five or fewer would be closer to 51%.
Or a statistician. Do you want to talk mathematical language or common parlance? The meanings are different. I assumed since you were discussing standard deviations, etc., that we were using the mathematical definitions.
The mode has no role in calculating expected value. Expectation is relevant, but certainly not the only important information.
So, is you response “No” or “Exactly”?
Obviously.
Gotcha. No problem. I don’t remember your original point, though.
Perhaps you have a definition with which I am not familiar. I googled the expression and found nothing to support you.
True, but I don’t see your point.
That is what we have both been saying. Where do we differ on this?
Yes really. $900 would be probabilistic breakeven; the other $100 would buy me peace of mind.
No. That peace of mind would be too expensive.
Thanks, I’ll remember that.
topdog on August 31, 2011 at 3:44 PM
The media (all networks) went WELL beyond reporting possibility of disaster. They reported impending doom. That’s what we’re complaining about.
Exactly.
I don’t apologize for speaking in realistic terms regarding neither the weather nor the economy. The models certainly aren’t built assuming perfect conditions.
That is 100% wrong.
Which IS the most likely outcome of an event.
1. Why should I? That mathematical description certainly dodesn’t conflict with the qualitative description of most likely.
2. If wikipedia was wrong, I wouldn’t dispute it – I would merely change it.
What the heck are you talking about? Have you been driven crazy?
A 3.5 is expected from rolling a die? Are you performing mathematical functions based on the number printed on the faces of the die? What if the die had letters on it instead of numbers? Would you claim that they expected outcome was C and a half? You’ve completely proven that you don’t know what you’re talking about.
Actually, there is no expected outcome from rolling a fair die once. Contrastingly, there is an expected outcome from rolling a successfully weighted die once. Also, there’s no expected outcome from flipping a coin once.
No, you’re not describing a confidence interval. If you’re using values along the distribution as the limits of your integral then you’re attempting to calculate the probability that the outcome will fall between those values.
What your describing here is the method for calculating the actually probability that the expected outcome will occur. This is NOT the method for calculating the expected outcome value.
Do you understand the difference between the expected outcome and the probability that the expected outcome will occur? In my 30 die roll example, the expected number of 6′s is five. Five is the expected outcome. The outcome distribution curve is NOT integrated in order to determine the value of the expected outcome.
Then tell me why you tried to claim “five or fewer” if it was so obviously wrong.
Are we in the Twilight Zone?
1. Your “five or fewer” claim is not in-line with the wikipedia definition.
2. Why on earth would someone hold a wikipedia definition in such high regard in such a debate? Are you trying to beclown yourself?
No, it’s not. It’s completely wrong because you’re using an completely incorrect approach.
Then you’re guess would be completely, hopelessly wrong.
Oh please. Don’t try to weasel your way out of this via a math vs. stats definition issue. You’re wrong using either definition.
Of course it does. In a data set, the mode, the most likely outcome, is the expected value.
You stated two different things. One was wrong and the other was exactly right.
Here’s what was exactly correct. You stated that the expected was most likely outcome and that most likely outcome didn’t need to have a probability greater than 50% in order to be the most likely outcome. An outcome of 10% can still be the most likely outcome and therefore be the expected outcome.
Actions based on prudent risk management would assume neutral risk tolerance. Actions are then biased from their baseline based on levels of risk tolerance. I don’t need any support from Google for this.
“Better safe than sorry” certainly doesn’t quantify the downside risk. “Better safe than sorry” leads one to allocate mitigation measures regardless of actual downside.
This thinking is EXACTLY the opposite of “better safe than sorry.”
“Better safe than sorry” thinkers would ask, “Why on earth wouldn’t you pay $25,000 in order to prevent the loss of $90,000???? That doesn’t make any sense.”
“Better safe than sorry” thinkers don’t think, “Better safe unless the cost of being safe is too expensive.”
blink on August 31, 2011 at 4:40 PM
Game over.
Thanks for playing
topdog on August 31, 2011 at 9:57 PM
I assume this means that you’ve finally learned something about statistics and risk management.
Here’s a tip: don’t accept everything on the internet as being accurate. That website is dead wrong for claiming that the expected outcome of a die roll is 3.5. In fact, it’s pure idiocy.
blink on September 1, 2011 at 8:10 AM
heh
topdog on September 1, 2011 at 8:51 AM
1/6+2/6+3/6+4/6+5/6+6/6=3.5
Pure Mathematics.
You won’t accept any authority higher than yourself, even the entire field of mathematics, because there is no authority higher than yourself. This is why you are no fun to play with.
Game over.
topdog on September 1, 2011 at 9:04 AM
That wasn’t very nice of me.
I don’t care to debate with you any longer, but I would be happy to explain why the expected value is 3.5. I know it is weird and counter-intuitive, but there is a simple explanation.
If I were to make you a gamble that if you pay me $3 I would pay you the amount that came up on one roll of a die, would that be a good bet? How about $4? The amount of a bet that would be fair to both of us turns out to be $3.50. That is the expected value.
If I roll a die 600 times, I would expect, on average, to get 100×1, 100×2, 100×3, …, 100×6. So the expected value of 600 rolls would be $2,100. That being the case, the expected value of each roll would be $2,100/600 = $3.50.
The word “expected” in statistics means something quite different than in everyday usage. While a roll of 3.5 is not only “unexpected” but virtually impossible, it is the “expected” value when speaking statistically.
The word “likely” does not mean the same thing as “expected” does in statistics. In fact, to the best of my knowledge, the word “likely” is not even defined as a statistical term. Sometimes “likely” is used incorrectly to describe the mode, but there is an important distinction that is often missed: the “mode” refers to actual data, whereas “likely” implies “probability”, which refers to as-yet undetermined outcomes.
I used to teach this stuff, but if you don’t want to take my word for it you can consult almost any textbook on algebra or probability or basic statistics. I will be happy to provide you internet-based references, but I suggest you do your own research. Try googling “expected value of one die roll”.
I hope this has been helpful.
topdog on September 1, 2011 at 10:10 AM
The arithmetic is correct, but the mathematics is wrong. 2+2=4 is correct arithmetic, but the mathematics is wrong if it’s your answer for finding the square of two.
Your performing the wrong function to find the expected outcome of a die roll.
The numbers on the face are insignificant. What is the face of the die were letters instead? What would the expected outcome be then?
Face it, you’re wrong.
You found this on some stupid website, and you label that an “authority higher” than me? Yes, that funny looking, cursive scripted website is wrong. Don’t believe everything you see on the internet.
No, the reason why I’m no fun for you to play with is because I always make you look like a stupid monkey.
blink on September 1, 2011 at 12:01 PM
Ha! I noticed that you were forced to multiply the number on the face of the die to get your answer the way you wanted it. Nice try.
Again, there is no expected outcome if a truly fair die is rolled. Each outcome has a 1/6 probability of being rolled. Therefore, there is no expected outcome.
Btw, using your method, what’s the expected outcome of a coin toss if the coin has “1″ printed on one side and “2″ printed on the other? Are you really going to claim “1.5″? This would truly show your stupidity.
I feel sorry for your students.
blink on September 1, 2011 at 12:05 PM
What I am about to write is entirely sincere, without irony or sarcasm.
You are obviously smarter than me because you fooled me for so long into thinking you were serious. You are smart enough to understand the concepts of normal distribution, sigma and skew, but not weighted averages? But not expected value? I don’t think so!
I lied earlier: it has been fun! I didn’t learn anything, but you certainly helped me to hone my thinking skills.
I am very intrigued by you and this game you are playing. Would you be willing to step out of character and chat? Or, is this sort of an Andy Kaufman thing that requires you to always stay in character?
Again, this is entirely sincere, without irony or sarcasm.
topdog on September 1, 2011 at 12:47 PM
Of course.
I think you mean expected outcome. Again, what’s the expected outcome of flipping a coin?
blink on September 1, 2011 at 2:30 PM
I get your point, and I think you get mine. You were using “outcome” and I was using “value”. If at any point I used the phrase “expected outcome” when I meant “expect value” I apologize.
As I said before, I thought we were talking statistics, and there “expected value” has a specific meaning. “Expected outcome” does not have a defined meaning in statistics. When an outcome is a value, then “expected outcome” is sometimes used synonymously with “expected value”, although I admit that that usage would be less than rigorous. When an outcome is NOT a value, then “expected outcome” is undefined statistically.
Statistically speaking, if the outcomes are heads and tails, there is no answer to your last question, since the phrase “expected outcome” is undefined. There are two possible outcomes, each equally probable. The outcomes have no value. There is no expected outcome.
In plain (non-statistical) language There are two possible outcomes, each equally likely. Neither outcome would be unexpected, neither would be expected. Note that as I have used the words here, “likely” and “expected” have different meanings.
If I recall correctly, that was the point you were disputing.
If instead the coins are labelled “1″ and “2″ the same answer applies. If one side is assigned the value of 1 and the other side the value of 2, then the expected value of the outcome would be 1.5. If your point is that it is imprecise to use the phrase “expected value of the outcome” in place of “expected outcome” then I accept your point.
Where are you going with this?
topdog on September 1, 2011 at 4:01 PM
On the other hand, I think you use “expected outcome” to mean “outcome with the highest probability”. I am not familiar with this usage in statistics.
topdog on September 1, 2011 at 4:14 PM
This is silly. I never stated that the “1″ and “2″ label were quantitative. Why do you say that there is no expected outcome if the coin is Heads and Tails, but there suddenly is an expected outcome as soon as you label the coins with different markings. You’re taking an excessively simplistic approach to this.
Also, if the Heads/Tails coin is weighted then there would be an expected outcome.
You might remember that we were discussing the possible outcome of the hurricane. You tried to claim that the term “most likely outcome” had to be an outcome with a greater than 50% probability – which you probably now realize is incorrect. A weighted die can have a only 25% chance of showing a (or “D” if they are letters) and be the expected outcome.
My point is that if the highest probability outcome is skewed towards less damage, then the media does everyone a disservice by frightening people with disproportionate coverage. An given thunderstorm has the potential for extensive damage – maybe you think the media should provide constant warnings prior to every thunderstorm.
Additionally, I’ve done more research about these hurricane models within the last week. Very little of their probability predications come from past data. Most of the predications are extrapolations of less powerful storm surges. I also think there’s quite a bit of ass-covering built into these predictions.
blink on September 3, 2011 at 7:01 PM
I am sorry you did not get my point. To much water under the bridge to try to recreate it.
That would have been ignorant if that had been me. I would have fervently argued the opposite. Maybe somebody else?
What I did say was ““Expected” implies a 50% cumulative probability” For example, if the expected wind speed at a certain point and time is 50 knots, then there is a 50% chance that the wind speed will be less than or equal to 50%. I also differentiated between “expected” and “likely”. Entire quote:
You have not dissuaded me of that position.
Can’t disagree. Disproportionate coverage would be, well, disproportionate.
Interesting observations. Thank you for passing them on.
topdog on September 6, 2011 at 5:25 PM
Correction:
What I did say was ““Expected” implies a 50% cumulative probability” For example, if the expected wind speed at a certain point and time is 50 knots, then there is a 50% chance that the wind speed will be less than or equal to
50%50 knots.topdog on September 6, 2011 at 5:40 PM
>>That would have been ignorant if that had been me. I would have fervently argued the opposite. Maybe somebody else?<<
No, it was you. You freaked out at the word "likely." You insisted that anything with the word likely had to mean greater than 50%.
Whatever. Your paragraph has nothing to do with assessing level of damage risk. A skewed probability distribution means that there’s not an equal probability that the wind velocity will 40 knots and an equal probability that the wind will be 60 knots. It’s quite possible that there’s a much higher probability for the winds to be 40 knots. This means that it’s quite possible that there is NOT an equal probability that the winds will be more damaging and an equal probability that the winds will be less damaging. In fact, I’m learning that the hurricane models (which are in their infancy stage) predictions are skewed to to energy loss in more northern latitudes.
Overall, your point didn’t address the issue which was being discussed.
blink on September 7, 2011 at 1:14 AM
Pay attention. I have never thought this to be true and never stated this to be true.
You could easily prove your point if it were true. It is not.
CITE ME!!!!!!!! (unintentional play on words!)
True. That follows from the definition of “skewed”.
Theoretically possible, of course. “quite” and “much” are subjective terms. If you are asserting that the probability distribution is skewed, I agree!
True, to the extent that the probability distribution is skewed. Just as the median can differ from the mean. I think I made it clear I was using a normal approximation (if not, my apologies…) Do you have any info suggesting that the skew is sufficient enough to make the normal distribution unusable as an approximation? Do you know whether the NHC models use a normal approximation? I do not know.
I think you mean “biased”? Interesting. I am surprised that water temperature is not adequately accounted for. That seems pretty basic… or is there some other factor in northern latitudes that they are not accounting for? (i am not challenging you…just curious for more on this).
We have both been OT for a while. My original point was that one cannot conclude that the media coverage was over-hyped solely on the basis that the actual damage was less than what was reported as reasonably possible. Also, a threat does not have to have a high probability (likelihood) to be worth preparing for. We don’t prepare for the possibility of being struck by a meteorite, but we do get out of the pool when we hear thunder even though actually being struck is not very likely.
Of course, I haven’t attempted to prove the opposite, that the media coverage was not over-hyped. In some cases it may have been, but that was not my experience. The NHC analyses I saw indicated a non-trivial possibility of disaster. I never heard any media source refer to disaster as anything more than a “possibility”. In my opinion, preparations were justified.
topdog on September 7, 2011 at 2:15 PM
You made that clear, and I repeatedly scoffed at that assumption. It’s much less likely that any distribution that occurs within nature to be normal rather than skewed.
Apparently, it has to do with the fact that hurricanes hitting the coast have crossed the Gulf Stream and as they continue further north they lose energy from the water and the land. Since most of the hurricane models use data from warmer water travels, and since much less data exists for actual hurricanes north of 35°N. Just look at Irene. There was no hurricane data collection north of 35°N since it was really just a tropical storm or even weaker.
Btw, I think it’s time for you to correct this statement. I’m quite sure you know that expected doesn’t mean a 50% probability for neither the expected value nor the expected outcome.
Think about it. There is not a 50% probability that a die roll will result in a 3.5.
Even if there’s a 50% probability of a value between 0 and 3.5 doesn’t mean that there’s a 50% probability of a 3.5.
blink on September 7, 2011 at 3:05 PM
And I repeatedly stated that my conclusion wasn’t based on this. Nothing I wrote indicated that it was.
And the hype didn’t report what was “reasonably” possible. The hype was reporting that much greater damage was probable.
I’ve been clear about this from my first comment. But most certainly it’s a waste of resources to prepare for a low probability threat which isn’t much of a threat. I’m quite happy debating what the probability of greater damage was and the extent of such greater damage, but you need to stop pretending that I’m arguing against the sophomoric statement that you just made.
Regardless of the actual level of risk of lighting striking the pool you’re in (which I’m quite happy to debate), the resources required to “getting out of a pool” is negligible. What the media hyped people into doing was not a negligible use of resources. “Better safe than sorry” implies that no consumption of resources in the name of preparation is imprudent. And you know that was my original point.
I didn’t see any predictions which justified the onslaught of hype.
I completely disagree. The preparations I witnessed were ridiculous.
blink on September 7, 2011 at 3:22 PM
True, but for statistical purposes it is extremely common to use normal as an approximation.
For a non-skewed (symmetrical) continuous function this is true, if you insert the word CUMULATIVE before “probability”. “Expected” implies a 50% cumulative probability for a normal distribution, and I think for any symmetrical (non-skewed) continuous distribution as well.
You are right: it means that there is a 50% cumulative probability of 3.5. Please stop arguing against my statements with the word “cumulative” removed.
“Definition for cumulative probability:
In probability theory and statistics, the cumulative distribution function (CDF), or just distribution function, describes the probability that a real-valued random variable X with a given probability distribution will be found at a value less than or equal to x.” – Wikipedia. Not good enough? Find a better definition.
I understood you to argue that the reported damage was far less than the hype. What was your point in mentioning this if not to bolster your claim that it was hype? (not a rhetorical question)
All right, I will take you at your word that the statement does not represent your intended argument. I am glad we agree that the statement is true. Now will you PLEASE quit arguing against my”cumulative probability” statements while mis-characterizing what I said by dropping the word “cumulative”.
The rest of what you wrote is simply our opinions and our experiences differing, and there is no further point in rehashing our positions.
topdog on September 7, 2011 at 5:00 PM
For the record, what you wrote was the following:
I interpreted this to mean: “The media was hyping disaster. Disaster did not happen. Therefore, the media hype was, in fact, hype.”
A reasonable interpretation, but I do understand that this is not what you meant to say. I am still in the dark regarding what you did mean by what you wrote.
topdog on September 7, 2011 at 10:08 PM
Only in academia. The real world uses actual data. The hurricane models are sh!t if they’re using normal distributions.
No, that’s fine. I’ll back off the cumulative probability issue.
Your poor understanding in the beginning was the majority of the problem here.
Go back and reread my original comment (this thread isn’t that long). It had nothing to do with the reported damaging being less than the hype.
My point in mentioning what? What is “this”? Again, go back and read my original comment.
Any comments I made about the actual hurricane damage were separate from my original comments regarding “better safe than sorry” being a terrible risk-management guide.
I can make two independent statements all day long. I can say that the sky is blue and that the milk is cold. One statement doesn’t need to be dependent on the other. You will not find any evidence of my conflating my two, independent points.
That’s an unreasonable interpretation.
Again, the statement was in response to someone that claimed that the hurricane damage was actually in-line with the media’s worse case scenario warnings. Obviously, it wasn’t.
We can argue about wether or not the media hyped the worse case scenario warnings or not. But there’s no way you can argue that the worse case scenarios occurred.
blink on September 9, 2011 at 2:36 AM
Thank you for your thoughts.
I’ll leave you with the last word.
See you on the next thread.
Over and out.
topdog on September 9, 2011 at 2:26 PM
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