Discussion
Suppose the distribution of a random variable is normal \((\mu, \sigma^{2})\). Then, which of the following statements is true ?
Answer: D
The true statement is the mean deviation from the mean is the same as the mean deviation from the median.
A normal distribution is symmetric, so its mean, median and mode coincide at \(\mu\). Mean deviation about the mean and about the median are then the same quantity, \(E|X - \mu| = \sigma\sqrt{2/\pi} \approx 0.8\sigma\).
The other statements have the standard values swapped or wrong. Symmetry makes the skewness \(\beta_{1} = \mu_{3}^{2}/\mu_{2}^{3} = 0\), not 3. The kurtosis \(\beta_{2} = \mu_{4}/\mu_{2}^{2} = 3\) (the normal is mesokurtic); it is the excess kurtosis \(\beta_{2} - 3\) that is 0. And \(\mu \pm \sigma\) holds about 68.27% of the area; 95% needs about \(\mu \pm 1.96\sigma\).
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