A state park has a normally distributed daytime temperatures in the month of June. The mean temperature is 75 Fahrenheit and then standard deviation of 7. What is the probability the temperature is above 82F?

0.16

Step-by-step explanation:

This probability is the area under the standard normal curve to the right of 82.  Note that 82 is ONE standard deviation above the mean:  75 + 7 = 82.

According to the Empirical Rule, 68% of data lies within one standard deviation of the mean.  In other words, 34% lies between the mean and one standard deviation above the mean.  The remaining area under the standard normal curve is thus (100% - 68%), or 32%.  Half of that is 16%.  

The probability that the temperature is above 82, or above 1 standard deviation above the mean, is 16%, which we can also express as 0.16.

Alternatively, use a calculator with built-in statistical functions.  In this case the desired result is found by executing the following command:

normcdf(82, 10000, 75, 7).  The desired result is 0.1587, or approximately 0.16.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.