Solved on Feb 08, 2024

Find the mean and standard deviation of $Y=4X-5$ when $X$ has mean $=29$ and standard deviation $=6$ .

STEP 1

Assumptions
1. The mean of the random variable $X$ is given as $\mu_X = 29$ .
2. The standard deviation of the random variable $X$ is given as $\sigma_X = 6$ .
3. The new random variable $Y$ is defined as a linear transformation of $X$ : $Y = 4X - 5$ .
4. The properties of linear transformations of random variables will be used to find the mean and standard deviation of $Y$ .

STEP 2

To find the mean of $Y$ , denoted as $E[Y]$ , we use the linearity of expectation, which states that for any constants $a$ and $b$ and a random variable $X$ :
$E[aX + b] = aE[X] + b$

STEP 3

Apply the formula to find $E[Y]$ using the given transformation $Y = 4X - 5$ :
$E[Y] = E[4X - 5]$

STEP 4

Substitute the known values of $E[X]$ and the constants into the formula:
$E[Y] = 4E[X] - 5$
$E[Y] = 4 \cdot 29 - 5$

STEP 5

Calculate the mean of $Y$ :
$E[Y] = 4 \cdot 29 - 5 = 116 - 5 = 111$

STEP 6

To find the standard deviation of $Y$ , denoted as $\sigma_Y$ , we use the property that for any constant $a$ and a random variable $X$ :
$\sigma[aX + b] = |a| \cdot \sigma[X]$

STEP 7

Apply the formula to find $\sigma_Y$ using the given transformation $Y = 4X - 5$ :
$\sigma_Y = \sigma[4X - 5]$

STEP 8

Since the addition of a constant does not affect the standard deviation, we only consider the multiplication by the constant:
$\sigma_Y = |4| \cdot \sigma_X$

STEP 9

Substitute the known value of $\sigma_X$ into the formula:
$\sigma_Y = 4 \cdot 6$

STEP 10

Calculate the standard deviation of $Y$ :
$\sigma_Y = 4 \cdot 6 = 24$

STEP 11

Combine the results for the mean and standard deviation of $Y$ :
The mean of $Y$ is $E[Y] = 111$ and the standard deviation of $Y$ is $\sigma_Y = 24$ .
The correct answer is:
$E[Y]=111 ; \sigma_{Y}=24$

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Find the mean and standard deviation of $Y=4X-5$ when $X$ has mean $=29$ and standard deviation $=6$ .

STEP 1

STEP 2

To find the mean of $Y$ , denoted as $E[Y]$ , we use the linearity of expectation, which states that for any constants $a$ and $b$ and a random variable $X$ :
$E[aX + b] = aE[X] + b$

STEP 3

Apply the formula to find $E[Y]$ using the given transformation $Y = 4X - 5$ :
$E[Y] = E[4X - 5]$

STEP 4

Substitute the known values of $E[X]$ and the constants into the formula:
$E[Y] = 4E[X] - 5$
$E[Y] = 4 \cdot 29 - 5$

STEP 5

Calculate the mean of $Y$ :
$E[Y] = 4 \cdot 29 - 5 = 116 - 5 = 111$

STEP 6

To find the standard deviation of $Y$ , denoted as $\sigma_Y$ , we use the property that for any constant $a$ and a random variable $X$ :
$\sigma[aX + b] = |a| \cdot \sigma[X]$

STEP 7

Apply the formula to find $\sigma_Y$ using the given transformation $Y = 4X - 5$ :
$\sigma_Y = \sigma[4X - 5]$

STEP 8

Since the addition of a constant does not affect the standard deviation, we only consider the multiplication by the constant:
$\sigma_Y = |4| \cdot \sigma_X$

STEP 9

Substitute the known value of $\sigma_X$ into the formula:
$\sigma_Y = 4 \cdot 6$

STEP 10

Calculate the standard deviation of $Y$ :
$\sigma_Y = 4 \cdot 6 = 24$

STEP 11

Combine the results for the mean and standard deviation of $Y$ :
The mean of $Y$ is $E[Y] = 111$ and the standard deviation of $Y$ is $\sigma_Y = 24$ .
The correct answer is:
$E[Y]=111 ; \sigma_{Y}=24$

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Find the mean and standard deviation of Y=4X−5Y=4X-5Y=4X−5 when XXX has mean =29=29=29 and standard deviation =6=6=6.

STEP 1

STEP 2

To find the mean of YYY, denoted as E[Y]E[Y]E[Y], we use the linearity of expectation, which states that for any constants aaa and bbb and a random variable XXX:E[aX+b]=aE[X]+bE[aX + b] = aE[X] + bE[aX+b]=aE[X]+b

STEP 3

Apply the formula to find E[Y]E[Y]E[Y] using the given transformation Y=4X−5Y = 4X - 5Y=4X−5:E[Y]=E[4X−5]E[Y] = E[4X - 5]E[Y]=E[4X−5]

STEP 4

Substitute the known values of E[X]E[X]E[X] and the constants into the formula:E[Y]=4E[X]−5E[Y] = 4E[X] - 5E[Y]=4E[X]−5E[Y]=4⋅29−5E[Y] = 4 \cdot 29 - 5E[Y]=4⋅29−5

STEP 5

Calculate the mean of YYY:E[Y]=4⋅29−5=116−5=111E[Y] = 4 \cdot 29 - 5 = 116 - 5 = 111E[Y]=4⋅29−5=116−5=111

STEP 6

To find the standard deviation of YYY, denoted as σY\sigma_YσY​, we use the property that for any constant aaa and a random variable XXX:σ[aX+b]=∣a∣⋅σ[X]\sigma[aX + b] = |a| \cdot \sigma[X]σ[aX+b]=∣a∣⋅σ[X]

STEP 7

Apply the formula to find σY\sigma_YσY​ using the given transformation Y=4X−5Y = 4X - 5Y=4X−5:σY=σ[4X−5]\sigma_Y = \sigma[4X - 5]σY​=σ[4X−5]

STEP 8

Since the addition of a constant does not affect the standard deviation, we only consider the multiplication by the constant:σY=∣4∣⋅σX\sigma_Y = |4| \cdot \sigma_XσY​=∣4∣⋅σX​

STEP 9

Substitute the known value of σX\sigma_XσX​ into the formula:σY=4⋅6\sigma_Y = 4 \cdot 6σY​=4⋅6

STEP 10

Calculate the standard deviation of YYY:σY=4⋅6=24\sigma_Y = 4 \cdot 6 = 24σY​=4⋅6=24

STEP 11

Combine the results for the mean and standard deviation of YYY:The mean of YYY is E[Y]=111E[Y] = 111E[Y]=111 and the standard deviation of YYY is σY=24\sigma_Y = 24σY​=24.The correct answer is:E[Y]=111;σY=24E[Y]=111 ; \sigma_{Y}=24E[Y]=111;σY​=24

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Find the mean and standard deviation of $Y=4X-5$ when $X$ has mean $=29$ and standard deviation $=6$ .

To find the mean of $Y$ , denoted as $E[Y]$ , we use the linearity of expectation, which states that for any constants $a$ and $b$ and a random variable $X$ :
$E[aX + b] = aE[X] + b$

Apply the formula to find $E[Y]$ using the given transformation $Y = 4X - 5$ :
$E[Y] = E[4X - 5]$

Substitute the known values of $E[X]$ and the constants into the formula:
$E[Y] = 4E[X] - 5$
$E[Y] = 4 \cdot 29 - 5$

Calculate the mean of $Y$ :
$E[Y] = 4 \cdot 29 - 5 = 116 - 5 = 111$

To find the standard deviation of $Y$ , denoted as $\sigma_Y$ , we use the property that for any constant $a$ and a random variable $X$ :
$\sigma[aX + b] = |a| \cdot \sigma[X]$

Apply the formula to find $\sigma_Y$ using the given transformation $Y = 4X - 5$ :
$\sigma_Y = \sigma[4X - 5]$

Since the addition of a constant does not affect the standard deviation, we only consider the multiplication by the constant:
$\sigma_Y = |4| \cdot \sigma_X$

Substitute the known value of $\sigma_X$ into the formula:
$\sigma_Y = 4 \cdot 6$

Calculate the standard deviation of $Y$ :
$\sigma_Y = 4 \cdot 6 = 24$

Combine the results for the mean and standard deviation of $Y$ :
The mean of $Y$ is $E[Y] = 111$ and the standard deviation of $Y$ is $\sigma_Y = 24$ .
The correct answer is:
$E[Y]=111 ; \sigma_{Y}=24$