Is a measure of 22 inches "far away" from a mean of 16 inches? As someone with knowledge of statistics, you answer "it depends" and request the standard deviation of the underlying data. (a) Suppose the data come from a sample whose standard deviation is 2 inches. How many standard deviations is 22 inches from 16 inches? (b) Is 22 inches far away from a mean of 16 inches? (c) Suppose the standard deviation of the underlying data is 4 inches. Is 22 inches far away from a mean of 16 inches?

a) 3 standard deviations above 16

b) More than 2 standard deviations of the mean, so yes, 22 inches is faw away from the mean of 16 inches.

c) Less than 2 standard deviations, so not far away.

Step-by-step explanation:

Z-score:

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If Z < -2 or Z > 2, X is considered to be far away from the mean.

In this question, we have that:

(a) Suppose the data come from a sample whose standard deviation is 2 inches. How many standard deviations is 22 inches from 16 inches?

This is Z when .

So

So 22 inches is 3 standard deviations fro 16 inches.

(b) Is 22 inches far away from a mean of 16 inches?

3 standard deviations, more than two, so yes, 22 inches is far away from a mean of 16 inches.

(c) Suppose the standard deviation of the underlying data is 4 inches. Is 22 inches far away from a mean of 16 inches?

Now

1.5 standard deviations from the mean, so 22 inches is not far away from the mean.

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.

y/x of point b= y/x of point a

step-by-step explanation:

(i got the same question, this was the correct answer.)