Solved on Feb 08, 2024

Find the mean and standard deviation of Y=4X5Y=4X-5 when XX has mean =29=29 and standard deviation =6=6.

STEP 1

Assumptions
1. The mean of the random variable XX is given as μX=29\mu_X = 29.
2. The standard deviation of the random variable XX is given as σX=6\sigma_X = 6.
3. The new random variable YY is defined as a linear transformation of XX: Y=4X5Y = 4X - 5.
4. The properties of linear transformations of random variables will be used to find the mean and standard deviation of YY.

STEP 2

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

STEP 3

Apply the formula to find E[Y]E[Y] using the given transformation Y=4X5Y = 4X - 5:
E[Y]=E[4X5]E[Y] = E[4X - 5]

STEP 4

Substitute the known values of E[X]E[X] and the constants into the formula:
E[Y]=4E[X]5E[Y] = 4E[X] - 5
E[Y]=4295E[Y] = 4 \cdot 29 - 5

STEP 5

Calculate the mean of YY:
E[Y]=4295=1165=111E[Y] = 4 \cdot 29 - 5 = 116 - 5 = 111

STEP 6

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

STEP 7

Apply the formula to find σY\sigma_Y using the given transformation Y=4X5Y = 4X - 5:
σY=σ[4X5]\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:
σY=4σX\sigma_Y = |4| \cdot \sigma_X

STEP 9

Substitute the known value of σX\sigma_X into the formula:
σY=46\sigma_Y = 4 \cdot 6

STEP 10

Calculate the standard deviation of YY:
σY=46=24\sigma_Y = 4 \cdot 6 = 24

STEP 11

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

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