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  1. Asked: April 7, 2021In: ,

    sampling distribution

    charizvalerie14
    charizvalerie14 Tyro
    This answer was edited.

    Please open the attached file.

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  2. Asked: April 6, 2021In: ,

    How to get Mean and Variance?

    charizvalerie14
    charizvalerie14 Tyro
    This answer was edited.

    7-8. Let, μ= ∑[x · P(X)] be your formula finding the Mean.  y 0 1 2 P(Y) 0.12 0.63 0.25 y · P(Y) 0 o.63 0.5 We multiplied y and P(Y) so, the products are 0, 0.63, 0.5 Then, we will just summate 0, 0.63, o.5 to find the Mean. 0 + 0.63 + o.5= 1.13 Therefore, the Mean is μ= 1.13 To find the Variance. LRead more

    7-8. Let, μ= ∑[x · P(X)] be your formula finding the Mean. 

    y 0 1 2
    P(Y) 0.12 0.63 0.25
    y · P(Y) 0 o.63 0.5

    We multiplied y and P(Y) so, the products are 0, 0.63, 0.5

    Then, we will just summate 0, 0.63, o.5 to find the Mean.

    0 + 0.63 + o.5= 1.13

    Therefore, the Mean is μ= 1.13

    To find the Variance. Let, σ2= ∑[x2 · P(X)]- μ2 be your formula finding the Variance.

    y P(Y) y2 y2· P(Y)
    0 0.12 0 0
    1 0.63 1 0.63
    2 0.25 4 1

    [0 + 0.63 + 1]- 1.132= 0.35

    Therefore, the Variance is σ2= 0.35

     

    9-10. Let, μ= ∑[x · P(X)] be your formula finding the Mean. 

    z 1 2 3
    P(Z) 0.23 0.45 0.32
    z · P(Z) 0.23 0.9 0.96

    0.23 + 0.9 + 0.96= 2.09

    Therefore, the Mean is μ= 2.09

    To find the Variance. Let, σ2= ∑[x2 · P(X)]- μ2 be your formula finding the Variance.

    z P(Z) z2 z2 · P(Z)
    1 0.23 1 0.23
    2 0.45 4 1.8
    3 0.32 9 2.88

    [0.23 + 1.8 + 2.88]- 2.092= 0. 54

    Therefore, the Variance is σ2= 0.54

     

     

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