If the variance of a variable x is 25, what is the standard deviation of y if y = 9 - 2x?
Understand the Problem
The question is asking to find the standard deviation of the variable y, which is defined as a linear transformation of another variable x. We use the relationship between variance and standard deviation, as well as how linear transformations affect variance, to solve it. Specifically, the standard deviation of y can be derived from the variance of x, taking into account the coefficient and constant in the linear equation.
Answer
The standard deviation of \(y\) is given by $\sigma_y = |a| \cdot \sigma_x$.
Answer for screen readers
The standard deviation of (y) is given by:
$$ \sigma_y = |a| \cdot \sigma_x $$
Steps to Solve
- Identify the linear transformation
Given that $y = ax + b$ where (a) is a coefficient and (b) is a constant, we know that $y$ is a linear transformation of $x$.
- Understand the relationship between variance and standard deviation
The variance $\sigma^2$ of a variable is related to the standard deviation $\sigma$ through the equation:
$$ \sigma = \sqrt{\sigma^2} $$
- Apply the rules for variance with linear transformations
For a linear transformation of the form (y = ax + b), the variance of (y) is given by:
$$ \text{Var}(y) = a^2 \cdot \text{Var}(x) $$
This is because adding a constant (b) does not affect the variance.
- Find the standard deviation of (y)
To find the standard deviation of (y), we apply the square root to the variance:
$$ \sigma_y = \sqrt{\text{Var}(y)} = \sqrt{a^2 \cdot \text{Var}(x)} $$
This simplifies to:
$$ \sigma_y = |a| \cdot \sigma_x $$
where $\sigma_x$ is the standard deviation of $x$.
The standard deviation of (y) is given by:
$$ \sigma_y = |a| \cdot \sigma_x $$
More Information
This formula shows how the standard deviation of (y) is directly affected by the coefficient (a) from the linear transformation and the standard deviation of (x). The absolute value of (a) is used because standard deviation cannot be negative.
Tips
- Ignoring the absolute value: Not using $|a|$ can lead to incorrect interpretations, especially if (a) is negative.
- Confusing variance with standard deviation: It's important to remember that standard deviation is the square root of variance.
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