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Peeved about a simple Sample Mean question

  1. Jul 20, 2013 #1
    Let X(bar)[sample mean] be the average of a sample of 16 independent normal random variables with mean 0 and variance 1. Determine c such that

    P(|X(bar)| < c) = 0.5


    Ok here's the problem, I know this is SUPPOSED to be a simple computation but for one reason or another little ol' fooliosh me can't figure it out.

    Since we are dealing with X(bar) I figure it is defined as 1/n[itex]\sum[/itex](from 1 to n) Xi

    now since we're talking abiout 16 independent Normal random variables, I figured it would be as easy as taking the sum of all of these normals and solving for x. Well according to the solution this isn't the case.

    Another thing I tried was taking the product of the iid normals:

    1/(2[itex]\pi[/itex][itex]\sigma2[/itex])1/2 e[itex]\sum[/itex](Xi2)/2

    set this equal to 0.5 and solve for X. Again the wrong answer. Help please
     
  2. jcsd
  3. Jul 20, 2013 #2

    jbunniii

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    If
    $$Y = \frac{1}{N}\sum_{n=1}^{N} X_n$$
    then what can you say about ##Y##? What kind of random variable is it? What are its mean and variance?
     
  4. Jul 20, 2013 #3

    HallsofIvy

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    You seem to be completely misunderstanding the problem! What you should be doing is using the theorem:

    "The average of n random variable, each having normal distribution with mean [itex]\mu[/itex] and standard deviation [itex]\sigma[/itex], has the standard deviation with mean [itex]]\mu[/itex] and standard deviation [itex]\frac{\sigma}{\sqrt{n}}[/itex]. "

    The problem is simply asking you to find c the probability of that "X" is less than c is 0.5 when X has the normal distribution with 0 and standard deviation 4. That can be done by looking up "z" such that P(z)< 0.5 on a table of the standard normal distribution
    (A good one online is at http://www.math.unb.ca/~knight/utility/NormTble.htm [Broken])
    and then solving [itex]\frac{x}{4}= z[/itex].

    (You really shouldn't need a table to find z such that P(z)< 0.5!)
    (I say after looking it up in the table myself!!:redface:)
     
    Last edited by a moderator: May 6, 2017
  5. Jul 20, 2013 #4

    Ok, I got the right solution, but I got it extremely flukey....Let me see if I make sense of it.

    So looking at the table of the standard normal distribution, for the value of 0.5 I got a probability of 0.6915. Taking this 0.6915 I put it into the expression of [itex]\frac{x}{4}= z[/itex] and got a value of 0.1728 for z. A couple questions: the 0.6915 is the percentage in relation to the z-score of 0.5? and what exactly was the motivation behind dividing 0.6915 by 4? Sure the answer might be right but what was that expression representing?
     
    Last edited by a moderator: May 6, 2017
  6. Jul 20, 2013 #5

    Ray Vickson

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    The sample mean ##\bar{X}## is normally distributed, has expected value = 0 and variance ##\sigma^2=16##, so has standard deviation ##\sigma = 4##. That is, we can write
    [tex] \bar{X} = \frac{Z}{\sigma} = \frac{Z}{4},[/tex]
    where ##Z \sim N(0,1).## So,
    [tex] P\{ |\bar{X}| < c \} = P\{ -c < \bar{X} < c \} = P\{-c < Z/4 < c\}.[/tex]
     
  7. Jul 20, 2013 #6

    Ray Vickson

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    The sample mean ##\bar{X}## is normally distributed, has expected value = 0 and variance ##\sigma^2=16##, so has standard deviation ##\sigma = 4##. That is, we can write
    [tex] \bar{X} = \sigma Z = 4 Z,[/tex]
    where ##Z \sim N(0,1).## So,
    [tex] P\{ |\bar{X}| < c \} = P\{ -c < \bar{X} < c \} = P\{-c < 4 Z < c\}.[/tex]
     
  8. Jul 20, 2013 #7

    ok, just to make sure this is a legitimate move I'm going to make we have established:
    [tex] P\{-c < 4 Z < c\}.[/tex] = 0.5

    now looking at the tables, for a Standard Normal distribution, the value of "c" is satisfied by 0.6915.

    Therefore: [tex] P\{-0.6915 < 4 Z < 0.6915\}.[/tex] = 0.5, but since this is not fully standardized I have to divide by the 4:

    therefore:

    [tex] P\{-0.6915/4 < Z < 0.6915/4\}.[/tex]

    which gives me 0.1728. Is this the right set of steps?
     
  9. Jul 20, 2013 #8

    Ray Vickson

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    Sorry: I meant that variance is 1/16 and ##\sigma = 1/4##. We have ##\bar{X} = \sigma Z = Z/4,## so the probability is ##P \{ -c < Z/4 < c \} = P \{ -4c < Z < 4c \}##.
     
    Last edited: Jul 20, 2013
  10. Jul 21, 2013 #9

    Ray Vickson

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    Your answer is 16 times too small.
     
  11. Jul 21, 2013 #10

    Ray Vickson

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    Please disregard this message, and instead refer to the others I gave. I wanted to delete this message, but the system won't allow it. Basically, ##\bar{X} = 4 Z##, NOT what I wrote immediately above.
     
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