Question:
Let \( f_1(x) \) be the probability density function of the \( N(0,1) \) distribution, and \( f_2(x) \) be the probability density function of the \( N(0, 6) \) distribution. Let \( Y \) be a random variable with probability density function
\[f(x) = 0.6 f_1(x) + 0.4 f_2(x), \quad -\infty < x < \infty.\]
Then \( \text{Var}(Y) \) is equal to:
Let \( f_1(x) \) be the probability density function of the \( N(0,1) \) distribution, and \( f_2(x) \) be the probability density function of the \( N(0, 6) \) distribution. Let \( Y \) be a random variable with probability density function
\[f(x) = 0.6 f_1(x) + 0.4 f_2(x), \quad -\infty < x < \infty.\]
Then \( \text{Var}(Y) \) is equal to:
\[f(x) = 0.6 f_1(x) + 0.4 f_2(x), \quad -\infty < x < \infty.\]
Then \( \text{Var}(Y) \) is equal to:
Updated On: Oct 1, 2024
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The Correct Option is B
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The correct option is (B): 3
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