Normal Distribution given a Probability Inequality?

Question

Given $ Y \sim \mathcal{N}(0.3, 0.7) $. Calculate $P(|Y − 0.2| < 0.8)$.

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This looks fairly simple, but I'm confused on which formula to use. I know that $$P(|X − \mu| < c) \le \frac{\operatorname{Var}(X)}{c^2}$$

But the $\mu$ in $ Y \sim \mathcal{N}(0.3, 0.7) $ is 0.3, while the $\mu$ in the inequality is 0.2. Does this matter? Since to find $\operatorname{Var}(X)$ i would just have to take square 0.7 from $ Y \sim \mathcal{N}(0.3, 0.7) $ ? Is $\frac{\operatorname{Var}(X)}{c^2}$ the solution or should I go onto trying to find a Normal Distribution?

Answer 1

Hints:

1. $P(|Y-a| < b) = P(a - b < Y < a + b)$ and
2. $Z = (Y-\mu) / \sigma$ is standard normal where $\mu = 0.3$ and $\sigma = \sqrt{0.7}$.