Find the standard deviation and coefficient of variation for $1, 9, 8, 7, 5$.

**Given data:** 1, 9, 8, 7, 5; $\displaystyle n = 5$ **Step 1: Calculate Mean.** $$\bar{x}=\frac{1+9+8+7+5}{5}=\frac{30}{5}=6$$$\displaystyle \bar{x} = \frac{1+9+8+7+5}{5} = \frac{30}{5} = 6$ **Step 2: Calculate deviations squared.** $$\overline{)\begin{array}{ccc}{x}_i& (x_i-\bar{x})& (x_i-\bar{x}{)}^2\\ 1& -5& 25\\ 9& 3& 9\\ 8& 2& 4\\ 7& 1& 1\\ 5& -1& 1\\ & & \sum =40\end{array}}$$$$\begin{array}{|c|c|c|} \hline x_i & (x_i -$$$\displaystyle \bar{x}$) & (x_i - $\displaystyle \bar{x}$)^2 \\ \hline 1 & -5 & 25 \\ 9 & 3 & 9 \\ 8 & 2 & 4 \\ 7 & 1 & 1 \\ 5 & -1 & 1 \\ \hline & & $\displaystyle \sum = 40$ \\ \hline \end{array} **Step 3: Calculate Variance.** $${\sigma }^2=\frac{\sum (x_i-\bar{x}{)}^2}{n}=\frac{40}{5}=8$$\displaystyle \sigma^2 = \frac{$\displaystyle \sum$(x_i - $\displaystyle \bar{x}$)^2}{n} = $\displaystyle \frac{40}{5} = 8$ **Step 4: Calculate SD.** $$\sigma =\sqrt{8}=2\sqrt{2}\approx 2.828$$$\displaystyle \sigma = \sqrt{8} = 2$$\displaystyle \sqrt{2}$ $\displaystyle \approx$ 2.828 **Step 5: Calculate CV.** $$CV =$$$\displaystyle \frac{\sigma}{\bar{x}}$ $\displaystyle \times$ 100 = $\displaystyle \frac{2.828}{6}$ $\displaystyle \times$ 100 $\displaystyle \approx$ 47.13\% Wait — let me recheck: $2.828 / 6 = 0.4713$\displaystyle , so$CV $\displaystyle \approx$ 47.13\%$. However, if we use the exact value $\displaystyle$$\displaystyle \sigma = \sqrt{8} = 2$.8284...: $$CV =$$$\displaystyle \frac{2.8284}{6}$ $\displaystyle \times$ 100 = 47.14\% The closest option to this computation is **$2.828, 47.13$** which is Option C. But the given `correct_option` is 'a' ($2.828, 49.32$). Let me verify $49.32$: $49.32\% \Rightarrow $\displaystyle \sigma = 49$.32$\displaystyle \times$ 6/100 = 2.959$, which doesn't match $$\displaystyle \sigma = 2$.828$. By direct computation, CV = $\displaystyle$$\displaystyle \frac{2.828}{6}$ $\displaystyle \times$ 100 $\displaystyle \approx$ 47.13\%. So Option C ($2.828, 47.13$) is mathematically correct. Hence, **Option C** is the correct answer.