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Answer: The resulting distribution will not be normal. In fact, it will be very heavily tailed.
**Explanation:** When mixing two normal distributions with different variances, the resulting mixture distribution is **not** normal. Specifically: - **Option A**: Incorrect - The mixture is not normal - **Option B**: Incorrect - With means both at 0, the mixture is unimodal, not bimodal - **Option C**: Correct - The mixture distribution exhibits heavy tails (leptokurtic) compared to a normal distribution - **Option D**: Incorrect - The mixture is symmetric (not skewed) since both component distributions have the same mean The mixture of normals with different variances creates a distribution with excess kurtosis, making it more peaked and heavy-tailed than a normal distribution.
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Consider two normal distributions, both with a mean of 0 but have different variances: N(0, 1) and N(0, 3). If we take the standard procedure to mix them, which of the following correctly describes the resulting mixture distribution?
A
The resulting distribution will also be normal, with a mean of 0 and a variance of 2.
B
The resulting distribution will not be normal. In fact, it will be bimodal.
C
The resulting distribution will not be normal. In fact, it will be very heavily tailed.
D
The resulting distribution will not be normal. In fact, it will be skewed and heavily tailed.
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