Temperate Summer In NYC..

[Mitchell]

Two things.

I want to make clear that I am NOT saying this to "brag" to anyone who is suffering through a blazing hot summer, or for those in areas where summer is usually brutal, but, except for.. a brief period.,. it really seems that this summer.. is rather.. temperate.

Today;s high in NYC is 88 degrees, about the warmest temp we've had recently. Storms are expected tonight, and then it is supposed to drop back into the high 70s to low 80s, with sunny skies. This is almost.. cool.. for late July.

Whereas I can remember other summers.. with long periods of days on end in the 90s, and even 100s, this summer doesn't seem to have such.

Is it just me.. or does this summer seem.. temperate?

 

[milagros317]

This is the coolest summer I can remember since moving back here in 1978. 😀
It follows the coldest winter I can remember since moving back here in 1978. 🙁
Climate alarmists will have to look elsewhere.

 

[Mitchell]

The cool weather continues. Highs this week are again going to be only in the upper 70s, or low 80s.

This feels downright wonderful. It's awesome not to have to deal with temps of 90 or above.

 

[milagros317]

Title:
Theorem for the coolest summer here in the last 40 years

Notation:
Where E is any event in a sample space, let P(E) represnt the
probability of E; let ~E represent the complementary event, that
E does not happen; and let P(~E) represent the probability of ~E.

Where E and F are events in a sample space and where F is possible,
let P(E|F) represent the conditional probabilty of E, given that F has
already happened.

THEOREM
Let A and B be events in a sample space. Let B be possible but not certain.
Then:
If P(A|B) > P(A), then P(A|~B) < P(A).

Statement of the theorem in words: If B happening makes A more probable,
then B not happening makes A less probable.

Proof of the theorem:

Let P(A) = a and P(B) = b. Note P(~B) = 1-b.
Let P(A and B) = x.

Recall that P(A|B) = x/b and we are given that this is greater than a.
Therefore, x > ab
Let d = x-ab. We know that d>0.

Note that P(A and ~B) = a-x = a-ab-d

Now note that P(A|~B) = P(A and ~B)/P(~B) = (a-ab-d)/(1-b)
= (a-ab)/(1-b) - d/(1-b) = a - d/(1-b) < a = P(A)
where the inequality is true because both d>0 and 1-b>0 (because B is not
certain).

quod erat demonstrandum



Punchline:
Let A be "Anthropogenic global warming is a real problem."
Let B be "This summer is warmer than normal."