Results 1 to 2 of 2

Math Help - Expected values and variance

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    96

    Expected values and variance

    So, the problem is:
    X is a random variable with E(x) = 100 and V(x) = 15. Find:
    a) E(x^2)
    b) E(3x+10)
    c) E(-x)
    d) V(-x)
    e) std dev(-x)


    For part a), i used the equation that V(x) = E(x^2) - (E(x))^2, so 15 = E(x^2) - 10000, therefore E(x^2) = 10015

    I'm not sure at all where to start with b, any pointers?

    for c, E(-x) would equal -100, right? I think this is assuming that the pmf still exists for x having negative values, and that the probabilities are the same...is this correct?

    using the same equation as part a, i found V(-x) to equal 15 as well, since squaring a negative number still results in a positive one, and so the std. dev of (-x) would be sqrt(15)?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,313
    Thanks
    1291
    Quote Originally Posted by mistykz View Post
    So, the problem is:
    X is a random variable with E(x) = 100 and V(x) = 15. Find:
    a) E(x^2)
    b) E(3x+10)
    c) E(-x)
    d) V(-x)
    e) std dev(-x)


    For part a), i used the equation that V(x) = E(x^2) - (E(x))^2, so 15 = E(x^2) - 10000, therefore E(x^2) = 10015

    I'm not sure at all where to start with b, any pointers?
    Since that is linear, E(3x+10)= 3E(x)+ 10.

    for c, E(-x) would equal -100, right? I think this is assuming that the pmf still exists for x having negative values, and that the probabilities are the same...is this correct?
    Yes, E(-x) is -100. No, it is not necessary for x to have negative values for this to make sense. E(f(x))= \sum f(x)*p(x) where the sum (or integral for continuous distributions) is taken over all possible values of x For example, suppose x has value 1 with prob 1/2 and value 2 with prob 1/2. Then E(x)= p(1)(-1)+ p(2)(-2)= (1/2)(-1)+ (1/2)(-2)= -3/2.

    using the same equation as part a, i found V(-x) to equal 15 as well, since squaring a negative number still results in a positive one, and so the std. dev of (-x) would be sqrt(15)?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expected value and variance
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: March 9th 2010, 10:30 PM
  2. expected value and variance
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: November 17th 2009, 06:14 AM
  3. [SOLVED] Finding expected values and variance
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 23rd 2009, 05:17 PM
  4. Expected Value and Variance Again
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 7th 2008, 01:10 PM
  5. Laws of Expected Values and Variance
    Posted in the Statistics Forum
    Replies: 1
    Last Post: February 12th 2008, 12:03 AM

Search Tags


/mathhelpforum @mathhelpforum