Let's evaluate each statement:
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Statement: Let \( X \) be a random variable that takes values between 0 and \( c \) only, for some \( c \geq 0 \) so that \( P(0 \leq X \leq c) = 1 \). Then, \( \text{Var}(X) \leq \frac{c^2}{4} \).
True or False?: True.
Explanation: The variance of a random variable is maximized when the distribution is uniform over the interval of possible values. If \( X \) is uniformly distributed over the interval \([0, c]\), the variance is given by:
\[ \text{Var}(X) = \frac{(b - a)^2}{12} = \frac{(c - 0)^2}{12} = \frac{c^2}{12} \]
Since \( \frac{c^2}{12} \leq \frac{c^2}{4} \) for any \( c \geq 0 \), the statement is true. Generally, for any random variable constrained to \([0, c]\), the variance will not exceed \(\frac{c^2}{4}\).
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Statement: Let \( X \) and \( Y \) be continuous random variables. If \( X \sim N(\mu, \sigma^2) \) (i.e., normal with mean \( \mu \) and variance \( \sigma^2 \)), \( Y = aX + b \) and \( a > 0 \), then \( Y \sim N(a\mu + b, a^2\sigma^2) \).
True or False?: True.
Explanation: If \( X \) is a normally distributed random variable, linear transformations of \( X \) (in the form \( Y = aX + b \)) will also be normally distributed. Specifically, when multiplying by a positive constant \( a \) and adding a constant \( b \), the following holds:
- The mean of \( Y \) is \( E[Y] = E[aX + b] = aE[X] + b = a\mu + b \).
- The variance of \( Y \) is \( \text{Var}(Y) = \text{Var}(aX + b) = a^2\text{Var}(X) = a^2\sigma^2 \).
Thus, \( Y \) is normally distributed with the parameters \( (a\mu + b, a^2\sigma^2) \), confirming the truth of the statement.
In summary:
- True
- True
1 answer