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Answer:

The mean and standard devaition of W are 6 and 31 respectively.

Step-by-step explanation:

Given that:

W = 7 + 2X

Variance (W) = Variance(7 + 2X)

For a term constant term, the variance is zero.

So;

[tex]W = Var (2X)[/tex]

W = 4 × Var(X)

W = 4 × (SD (X))²

W = 4 × (3)²

W = 4 × 9

W = 36

Thus, Variance (W) = 36

The standard deviation of W = [tex]\sqrt{variance}[/tex]

W = [tex]\sqrt{36}[/tex]

W = 6

Also;

E(W) = E(7 +2 X)

E(W) = 7 + 2 × E(X)

where;

mean X = 12

E(W) = 7 + 2(12)

E(W) = 31

Thus, the mean and standard devaition of W are 6 and 31 respectively.

The mean and the standard deviation of W are 31 and 6 respectively.

The given parameters are:

[tex]\mathbf{\bar x_x = 12}[/tex]

[tex]\mathbf{\sigma_x = 3}[/tex]

The definition of W is given as:

[tex]\mathbf{W = 7 + 2X}[/tex]

This means that:

[tex]\mathbf{Var(W) = Var(7) + Var(2X)}[/tex]

So, we have:

[tex]\mathbf{Var(W) = 0 + Var(2X)}[/tex] --- the variance of a constant term is 0

[tex]\mathbf{Var(W) = Var(2X)}[/tex]

This gives

[tex]\mathbf{Var(W) = 2^2 \times Var(X)}[/tex]

We have:

[tex]\mathbf{Var(X) = \sigma^2_x}[/tex]

So, the equation becomes

[tex]\mathbf{Var(W) = 2^2 \times \sigma^2_x}[/tex]

[tex]\mathbf{Var(W) = 2^2 \times 3^2}[/tex]

[tex]\mathbf{Var(W) = 36}[/tex]

The standard deviation is then calculated as:

[tex]\mathbf{\sigma_W = \sqrt{Var(W)}}[/tex]

[tex]\mathbf{\sigma_W = \sqrt{36}}[/tex]

[tex]\mathbf{\sigma_W = 6}[/tex]

The mean is then calculated as:

[tex]\mathbf{E(W) = E(7 + 2X)}[/tex]

This gives

[tex]\mathbf{E(W) = 7 + 2E(X)}[/tex]

So, we have:

[tex]\mathbf{E(W) = 7 + 2 \times 12}[/tex]

[tex]\mathbf{E(W) = 31}[/tex]

Hence, the mean and the standard deviation of W are 31 and 6 respectively.

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