Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Probability and gauss integral

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data

    1) Find the area under the curve for:
    a) P( 0 <= Z <= 2.07)
    b) P(-.64 <= Z < -.11)
    c) P( Z > -1.06)
    d) P(Z < -2.33)
    e) P(Z >= 4.61)

    2) a) Evaluate integral from 0 to 1.24 of e^(-x^2/2)
    b) Evaluate the integral from -inf to inf of 6*e^(-x^2/2)

    2. Relevant equations

    Z value table
    Power series expansion of e^x

    3. The attempt at a solution

    1)
    a) Fz(2.07) - Fz(0) = .9808 - .5 = .4808
    b) Fz(-.64) - Fz(a) = .4247 - .2611 = .163
    c) 1 - Fz(-1.06) = .85
    d) .0104
    e) .00002

    Can someone explain to me the difference in calculating an area where Z < x vs. Z <= x. Do I just use the value closest to x and less than it on the table for Z < x?
    2) a)I used the power series expansion and dervied each term upto n = 4 and got
    .6807
    b) No clue how to do this. Polar coordinates?
     
  2. jcsd
  3. Dec 19, 2008 #2

    statdad

    User Avatar
    Homework Helper

    With the standard normal distribution (indeed, any continuous probability distribution)
    there is no difference working with [tex] < [/tex] and [tex] \le [/tex], so that

    [tex]
    P(Z < z) = P(Z \le z)
    [/tex]

    As an example

    [tex]
    P(Z < 1.24) = P(Z \le 1.24)
    [/tex]

    Both can be found from a table or software.

    For the integral problem, consider this. If I use [tex] Z = 2.1[/tex]

    [tex] \begin{align*}
    P(Z \le 2.1) & = \int_{-\infty}^{2.1} \frac 1 {\sqrt{2 \pi}} e^{-{x^2}/2} \, dx\\
    & = \int_{-\infty}^0 \frac 1 {\sqrt{2 \pi}} e^{-{x^2}/2} \, dx + \int_0^{2.1} \frac 1 {\sqrt{2 \pi}} e^{-{x^2}/2} \, dx \\
    & = 0.5 + \frac 1 {\sqrt{2 \pi}} \int_0^{2.1} e^{-{x^2}/2} \, dx
    \end{align*}
    [/tex]

    If you rearrange terms you find that

    [tex]
    \int_0^{2.1} e^{-{x^2}/2} \, dx = \sqrt{2 \pi} \left(P(Z \le 2.1) - .5 \right)
    [/tex]

    None of the terms on the right require a series expansion.
    This idea should help you with your integral questions.
     
  4. Dec 19, 2008 #3
    Great explanations.

    a) = sqrt(2pi) * (P(Z <= 1.24) - .5)
    b) = sqrt(2pi)/6

    Is this right? Thanks.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook