1. **State the problem:** We have a random variable $X$ with probability mass function (pmf) $f(x) = \frac{1}{6}$ for $x = 1, 2, 3, 4, 5, 6$ and $0$ otherwise. We define a new random variable $Y = 5X - 2$. We want to find the standard deviation of $Y$. 2. **Recall formulas:** - The variance of $X$ is $\mathrm{Var}(X) = E[X^2] - (E[X])^2$. - For a linear transformation $Y = aX + b$, the variance is $\mathrm{Var}(Y) = a^2 \mathrm{Var}(X)$. - The standard deviation is $\sigma_Y = \sqrt{\mathrm{Var}(Y)}$. 3. **Calculate $E[X]$:** $$E[X] = \sum_x x f(x) = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{1}{6} \times 21 = 3.5$$ 4. **Calculate $E[X^2]$:** $$E[X^2] = \sum_x x^2 f(x) = \frac{1}{6}(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = \frac{1}{6}(1 + 4 + 9 + 16 + 25 + 36) = \frac{1}{6} \times 91 = 15.1667$$ 5. **Calculate $\mathrm{Var}(X)$:** $$\mathrm{Var}(X) = E[X^2] - (E[X])^2 = 15.1667 - (3.5)^2 = 15.1667 - 12.25 = 2.9167$$ 6. **Calculate $\mathrm{Var}(Y)$:** Since $Y = 5X - 2$, $a=5$, $b=-2$, so $$\mathrm{Var}(Y) = 5^2 \times \mathrm{Var}(X) = 25 \times 2.9167 = 72.9167$$ 7. **Calculate standard deviation $\sigma_Y$:** $$\sigma_Y = \sqrt{72.9167} \approx 8.54$$ **Final answer:** The standard deviation of $Y$ is approximately $8.54$.