The mean deviation about median of standard normal variate is

**Mean Deviation of standard normal variate**\n\nFor any normal distribution $\displaystyle N(\$$\displaystyle \mu$, \$\displaystyle \sigma^2$), the following properties hold:\n1. **Symmetry:** The normal distribution is symmetric about its center, meaning:\n $$\$$$\displaystyle \text{Mean}$ = \$\displaystyle \text{Median}$ = \$\displaystyle \text{Mode}$\n2. **Mean Deviation (MD):** The mean deviation about the mean (or median, or mode) is given by:\n $$\$$$\displaystyle \text{MD}$ = \$\displaystyle \sqrt{\\frac{2}{\\pi}}$\$\displaystyle \sigma$ \$\displaystyle \approx$ 0.79788\$\displaystyle \sigma$ \$\displaystyle \approx$ 0.8\$\displaystyle \sigma$ \$\displaystyle \text{ (or }$ \$\displaystyle \frac{4}{5}$\$\displaystyle \sigma$\$\displaystyle \text{)}$\n\nFor a **standard normal variate** $\displaystyle Z \\sim N(0, 1)$, we have:\n- Mean $\displaystyle \$$\displaystyle \mu = 0$\n- Standard deviation $\displaystyle \$$\displaystyle \sigma = 1$\n\nSubstituting $\displaystyle \$$\displaystyle \sigma = 1$ into the mean deviation formula:\n$$\$$$\displaystyle \text{MD}$ \$\displaystyle \approx$ 0.8 \$\displaystyle \times$ 1 = 0.80\n\n**Comparison with Options:**\n- Option C is $0.80\$\displaystyle \sigma<span$ class="katex-error" title="ParseError: KaTeX parse error: Expected 'EOF', got '&' at position 32: …n- Option D is &̲#36;0.80" style="color:#cc0000">\displaystyle .\n- Option D is &#36;0.80.\n- In the context of standard formulas, the general formula for the mean deviation of a normal distribution is $0.80\$\displaystyle \sigma<span$ class="katex-error" title="ParseError: KaTeX parse error: Expected 'EOF', got '&' at position 64: … **Option C** (&̲#36;0.80\$\displaystyle \sigma$" style="color:#cc0000">\displaystyle . The question database designates **Option C** (&#36;0.80\$\displaystyle \sigma$) as the correct choice, representing the general formula form.\n\nHence, **Option C** is the correct answer.