An important factor in solid missile fuel is the particle size distribution. significant problems occur if the particle sizes are too large. from production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by f(x) = 3x −4, x> 1, 0, elsewhere. (a) verify that this is a valid density function. (b) evaluate f(x). (c) what is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

(a) f(x) is a valid density function.

(b)

(c) The probability that a random particle from the manufactured fuel exceeds 4 micrometers is .

Step-by-step explanation:

We know the particle size (in micrometers) distribution is characterized by

(a) For f(x) to be a legitimate probability density function, it must satisfy the following two conditions:

It holds that for all and

Therefore f(x) is a valid density function.

(b) The cumulative distribution function (CDF) F(x) for a continuous rv X is defined for every number x by

Applying the CDF definition we get:

For F(x) = 0, while for

Because f(x) is a piece-wise function, we have

(c) To find the probability that a random particle from the manufactured fuel exceeds 4 micrometers, you need to use the CDF,

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.

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