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Math Help - uniform distribution of random independent variables

  1. #1
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    uniform distribution of random independent variables

    okay. my exam is approaching in a few days and I have one last question.

    Let X1, X2, ... X7 be a set of seven independent random variables, each
    having the uniform distribution of values from the [3, 6] interval (this is
    called random independent sample of size 7 from this distribution).

    (a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7.
    (b) What is the probability that more than three of these seven random
    variables will have a value smaller than 1
    3?
    (c) Find the skewness and kurtosis of X1.

    answers:

    (a) E(x) = 7(6+3)/2 = 31.5
    Var(x) = 7*[(6-3)^2]/2
    standard deviation = 2.2913

    (b) not sure

    (c) would skewness be 0 since its uniform? what about kurtosis
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  2. #2
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    Quote Originally Posted by chrisc View Post
    okay. my exam is approaching in a few days and I have one last question.

    Let X1, X2, ... X7 be a set of seven independent random variables, each
    having the uniform distribution of values from the [3, 6] interval (this is
    called random independent sample of size 7 from this distribution).

    (a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7.
    (b) What is the probability that more than three of these seven random
    variables will have a value smaller than 1
    3? Mr F says: The question is not clear to me. Is it meant to be 13, 1 or something else?
    (c) Find the skewness and kurtosis of X1.

    answers:

    (a) E(x) = 7(6+3)/2 = 31.5
    Var(x) = 7*[(6-3)^2]/2
    standard deviation = 2.2913

    (b) not sure

    (c) would skewness be 0 since its uniform? what about kurtosis
    (a) You're expected to know or derive that E(X_i) = \frac{6+3}{2} = \frac{9}{2} and Var(X_i) = \frac{(6-3)^2}{12} = \frac{3}{4}.

    Then:

    E\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} E(X_i) = 7 \cdot \frac{9}{2} = \frac{63}{2}.

    Var\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} Var(X_i) = 7 \cdot \frac{3}{4} = \frac{21}{4} (summing the variances follows because the X's are independent).


    (b) Is there a typo?


    (c) What definition/type of kurtosis are you using?
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    a) okay, confirmed

    b) was a typo, its 1/3, not 1

    c) nothing else was specified outside of this question
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    Quote Originally Posted by chrisc View Post
    a) okay, confirmed

    b) was a typo, its 1/3, not 1 Mr F says: Well if that's the case, what's the probability that any of the X's can be smaller than 1/3 .....?! Look at the interval over which the pdf is not equal to zero!

    c) nothing else was specified outside of this question
    c) That's why you go to your class notes or textbook, find the definition/type that's been used and then report the answer in this thread. I had thought that's what you'd do.
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    this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks)

    kurtosis:
    [E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ...
    [24β4−46β4+62β4−3β4]/β4 = 9

    does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam
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  6. #6
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    Quote Originally Posted by chrisc View Post
    this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks)

    kurtosis:
    [E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ...
    [24β4−46β4+62β4−3β4]/β4 = 9

    does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam
    Then calculate each of the bits in the formula:

    E(X^n) = \int_3^6 x^n \cdot 1 \, dx.

    \mu = E(X_1) = \frac{9}{2}.

    \sigma = s.d.(X_1) = \frac{\sqrt{3}}{2}.

    Now substitute into the formula.
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