Scores of an IQ test have a bell-shaped distribution with a mean of 100 and a standard deviation of 17. Use the empirical rule to determine the following. (a) What percentage of people has an IQ score between 66 and 134? (b) What percentage of people has an IQ score less than 83 or greater than 117? (c) What percentage of people has an IQ score greater than 117? (a) 95% (Type an integer or a decimal.) (b) (Type an integer or a decimal.)

(a) 95% of people have an IQ score between 66 and 134.

(b) 32% of people have an IQ score of less than 83 or greater than 117.

(c) 16% of people have an IQ score greater than 117.

Step-by-step explanation:

We are given that scores of an IQ test have a bell-shaped distribution with a mean of 100 and a standard deviation of 17.

Let X = scores of an IQ test

SO, X ~ Normal()

The z-score probability distribution for the normal distribution is given by;

                      Z  =    ~ N(0,1)

Now, the empirical rule states that;

(a) The percentage of people having an IQ score between 66 and 134 is given by;

     z score for 66 =    =    = -2

     z score for 134 =    =    = 2

This means that the percentage of people having an IQ score between 66 and 134 is 95%.

(b) Firstly, we have to find the percentage of people having an IQ score between 83 and 117;

     z score for 83 =    =    = -1

     z score for 117 =    =    = 1

This means that the percentage of people having an IQ score between 83 and 117 is 68%.

This means that the percentage of people having an IQ score of less than 83 or greater than 117 = 100% - 68% = 32%.

(c) In the above part, we find that there are 32% of the values that have an IQ score of less than 83 or greater than 117.

This means that half of this will be less than 83 and half of this will be greater than 117.

So, the percentage of people having an IQ score greater than 117 = = 16%.

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

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.