Math / Algebra

QuestionAssume $X$ has a normal distribution $N\left(9,5^{2}\right)$ . Find $E(5 X-4)^{2}$
Answer: $\square$

Studdy Solution

STEP 1

1. $X$ is a normally distributed random variable with mean $\mu = 9$ and variance $\sigma^2 = 25$ .
2. We need to find the expected value of the expression $(5X - 4)^2$ .
3. The properties of expectation and variance for linear transformations of random variables will be used.

STEP 2

1. Identify the transformation of the random variable.
2. Apply the properties of expectation.
3. Calculate the expected value.

STEP 3

Identify the transformation of the random variable $X$ . We are given the expression $(5X - 4)^2$ . This is a transformation of $X$ where $Y = 5X - 4$ .

STEP 4

Apply the properties of expectation. We need to find $E(Y^2)$ where $Y = 5X - 4$ . The expectation of a square of a linear transformation can be found using the formula:
$E(Y^2) = \text{Var}(Y) + [E(Y)]^2$

STEP 5

Calculate $E(Y)$ and $\text{Var}(Y)$ :
1. $E(Y) = E(5X - 4) = 5E(X) - 4 = 5 \times 9 - 4 = 45 - 4 = 41$
2. $\text{Var}(Y) = \text{Var}(5X - 4) = 5^2 \text{Var}(X) = 25 \times 25 = 625$

STEP 6

Now, substitute these values into the formula for $E(Y^2)$ :
$E(Y^2) = \text{Var}(Y) + [E(Y)]^2 = 625 + 41^2$
Calculate $41^2$ :
$41^2 = 1681$
Therefore:
$E(Y^2) = 625 + 1681 = 2306$
The expected value $E(5X - 4)^2$ is $\boxed{2306}$ .

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QuestionAssume $X$ has a normal distribution $N\left(9,5^{2}\right)$ . Find $E(5 X-4)^{2}$
Answer: $\square$

Studdy Solution

STEP 1

1. $X$ is a normally distributed random variable with mean $\mu = 9$ and variance $\sigma^2 = 25$ .
2. We need to find the expected value of the expression $(5X - 4)^2$ .
3. The properties of expectation and variance for linear transformations of random variables will be used.

STEP 2

1. Identify the transformation of the random variable.
2. Apply the properties of expectation.
3. Calculate the expected value.

STEP 3

Identify the transformation of the random variable $X$ . We are given the expression $(5X - 4)^2$ . This is a transformation of $X$ where $Y = 5X - 4$ .

STEP 4

Apply the properties of expectation. We need to find $E(Y^2)$ where $Y = 5X - 4$ . The expectation of a square of a linear transformation can be found using the formula:
$E(Y^2) = \text{Var}(Y) + [E(Y)]^2$

STEP 5

Calculate $E(Y)$ and $\text{Var}(Y)$ :
1. $E(Y) = E(5X - 4) = 5E(X) - 4 = 5 \times 9 - 4 = 45 - 4 = 41$
2. $\text{Var}(Y) = \text{Var}(5X - 4) = 5^2 \text{Var}(X) = 25 \times 25 = 625$

STEP 6

Now, substitute these values into the formula for $E(Y^2)$ :
$E(Y^2) = \text{Var}(Y) + [E(Y)]^2 = 625 + 41^2$
Calculate $41^2$ :
$41^2 = 1681$
Therefore:
$E(Y^2) = 625 + 1681 = 2306$
The expected value $E(5X - 4)^2$ is $\boxed{2306}$ .

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QuestionAssume XXX has a normal distribution N(9,52)N\left(9,5^{2}\right)N(9,52). Find E(5X−4)2E(5 X-4)^{2}E(5X−4)2Answer: □\square□

Studdy Solution

STEP 1

STEP 2

1. Identify the transformation of the random variable.2. Apply the properties of expectation.3. Calculate the expected value.

STEP 3

Identify the transformation of the random variable X X X. We are given the expression (5X−4)2 (5X - 4)^2 (5X−4)2. This is a transformation of X X X where Y=5X−4 Y = 5X - 4 Y=5X−4.

STEP 4

Apply the properties of expectation. We need to find E(Y2) E(Y^2) E(Y2) where Y=5X−4 Y = 5X - 4 Y=5X−4. The expectation of a square of a linear transformation can be found using the formula:E(Y2)=Var(Y)+[E(Y)]2 E(Y^2) = \text{Var}(Y) + [E(Y)]^2 E(Y2)=Var(Y)+[E(Y)]2

STEP 5

STEP 6

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QuestionAssume $X$ has a normal distribution $N\left(9,5^{2}\right)$ . Find $E(5 X-4)^{2}$
Answer: $\square$

1. Identify the transformation of the random variable.
2. Apply the properties of expectation.
3. Calculate the expected value.

Identify the transformation of the random variable $X$ . We are given the expression $(5X - 4)^2$ . This is a transformation of $X$ where $Y = 5X - 4$ .

Apply the properties of expectation. We need to find $E(Y^2)$ where $Y = 5X - 4$ . The expectation of a square of a linear transformation can be found using the formula:
$E(Y^2) = \text{Var}(Y) + [E(Y)]^2$