Probability of interval around normally distributed variable containing the mean

Question

Assume a normally distributed random variable X with mean m and standard deviation 1.2. We can say that ~95% of all samples of X lie in the interval [m - 2.4, m + 2.4].

I read in a book that this statement is equivalent to saying that for 95% of samples X, the interval [X - 2.4, X + 2.4] contains m.

How are the statements equivalent?

Answer 1

The argument is $$m-2.4 \le X \iff m \le X+2.4 \quad \text{ and } \quad X \le m+2.4 \iff X-2.4 \le m$$

so $$m -2.4 \le X \le m+2.4 \iff X -2.4 \le m \le X+2.4$$

and thus $$\mathbb P(m -2.4 \le X \le m+2.4) = \mathbb P(X -2.4 \le m \le X+2.4)$$

Answer 2

When $m-2.4\le X\le m+2.4$, then $-2.4\le X-m\le2.4$ or $-2.4\le m-X\le2.4$ and $X-2.4\le m\le X+2.4$.