Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)
-   -   Suppose that the probability distribution of a random variable x can be described by (https://www.askmehelpdesk.com/showthread.php?t=474117)

  • May 27, 2010, 07:32 PM
    azbrowneyedgirl
    Suppose that the probability distribution of a random variable x can be described by
    Suppose that the probability distribution of a random variable x can be described by the formula P(x) X/15

    For each of the values x =1, 2, 3, 4, 5. For example, then,
    P (x=2) =p (2) =2/15

    a) Write out the probability distribution of x.

    b) Show that the probability distribution of X satisfies the properties of a discrete probability distribution.

    c) Calculate the mean of x.

    d) Calculate the variance, and the standard deviation,
  • May 28, 2010, 09:18 AM
    Unknown008

    Note that we are not supposed to give you direct answers.

    For your first part, can you complete this table?



    This will be your answer to a)

    b) For this one, you need to check your notes, there should be somewhere saying how to do that.

    c) The mean of x () is given by , or

    d) Variance () is given by

    You already have (E(X))^2 from c), to find E(X^2), you need to do something similar to that of E(X), but you square the value of x like this:



    The standard deviation() is given by

    Post your answers! :)
  • Sep 18, 2010, 05:51 PM
    duffer987
    Suppose that the probability distribution of a random variable x can be described by the formula
    p(x) = x 15
    for each of the values x = 1, 2, 3, 4, and 5. For example, then, P(x = 2) = p(2) = 2=15.
    a. Write out the probability distribution of x.
    p(x) = x 15
    p(1) = 1/15
    p(2) = 2/15
    p(3) = 3/15
    p(4) = 4/15
    p(5) = 5/15
    b. Show that the probability distribution of x satisfies the properties of a discrete probability distribution.
    p(1) = 1/15 = .067
    p(2) = 2/15 = .133
    p(3) = 3/15 = .20
    p(4) = 4/15 = .267
    p(5) = 5/15 = .333
    c. Calculate the mean of x.
    µx = ∑ xp(x)
    µx = 1p(1) + 2p(2) + 3p(3) + 4p(4) + 5p(5)
    µx = 1(.067) + 2(.133) + 3(.20) + 4(.267) + 5(.333)
    µx = .067 + .266 + .60 + 1.068 + 1.665
    µx = 3.666
    d. Calculate the variance, s2x, and the standard deviation, sx.
    s2x
    o2 = ∑ (x - µx)2 p(x)
    = (1 – 3.666)2 p(1) + (2 – 3.666)2 p(2) + (3 – 3.666)2 p(3) + (4 – 3.666)2 p(4) + (5 – 3.666)2 p(5)
    = (-2.666)(.067) + (-1.666)(.133) + (-.666)(.20) + (.334)(.267) + (1.334)(.333)
    = -.178 + -.222 + -.133 + .089 + .444
    = 0
    sx
    o = √0
    o = 0
  • Sep 19, 2010, 06:11 AM
    Unknown008

    For b), you actually have to show that all the probabilities add up to 1. So, above poster is wrong.

    For d), Variance is given by:



    Meaning that the above poster is wrong again.

    And:

  • Sep 24, 2010, 12:51 PM
    duffer987
    Suppose that the probability distribution of a random variable x can be described by the formula
    p(x) = x 15
    for each of the values x = 1, 2, 3, 4, and 5. For example, then, P(x = 2) = p(2) = 2=15.
    a. Write out the probability distribution of x.
    p(x) = x 15
    p(1) = 1/15
    p(2) = 2/15
    p(3) = 3/15
    p(4) = 4/15
    p(5) = 5/15
    b. Show that the probability distribution of x satisfies the properties of a discrete probability distribution.
    p(1) + p(2) + p(3) + p(4) + p(5) = x

    (1 + 2 + 3 + 4 + 5)
    15

    x = 1
    c. Calculate the mean of x.
    µx = ∑ xp(x)
    µx = 1p(1) + 2p(2) + 3p(3) + 4p(4) + 5p(5)
    µx = 1(.067) + 2(.133) + 3(.20) + 4(.267) + 5(.333)
    µx = .067 + .266 + .60 + 1.068 + 1.665
    µx = 3.666
  • Sep 24, 2010, 01:01 PM
    Unknown008

    Well, you are not being really helpful reposting what you did earlier and correcting what I already said what was to be done... :rolleyes:

  • All times are GMT -7. The time now is 01:11 PM.