I'm learning the concept of confidence interval. One of the important inference is that "the probability that sample mean is within two standard deviation of population mean equals to 95%." is equivalent to "there is a 95% probability that population mean is within two standard deviation of sample mean."
It seems the logic is if $a-c \lt b \lt a+c$ then to $b-c \lt a \lt b+c$ where a = population mean, b = sample mean and c = 2SD.
I don't understand.
On
$$|\hat p - mean| < 2*(SD)$$ means "the distance from $\hat p$ to the mean (which is symmetric, signless) is less than 2 standard deviations". Using inequality rules:
$\hat p - mean < 2*(SD)$ or $\hat p - mean > -2 * (standard deviations)$
therefore
$\hat p < mean + 2*(SD)$ or $\hat p > mean - 2 * (SD)$
or
$b < a + c$ or $b > a - c$
, as you would write it (your first expression).
You can also collect $a$ to one side and see how it affects $b$ and $c$.
The compound inequalities mean the same thing when you write with ors. But you write your equations as if these ors were ands, and that's why you're confused.
Great to learn about confidence intervals by the way.
On
"$a-c<b<a+c$" is a placeholder for "$(a-c<b)\land (b<a+c)$", which in turn is equivalent to $(a<b+c)\land (b-c<a)$ by carrying the appropriate term to the other sides.
Let $\mu$ and $\bar{x}$ be the population and sample means, respectively. The confidence interval: $$\bar{x}-ME<\mu<\bar{x}+ME \ \ \ \ \ \ \ \ \ \ \ (1)$$ where $ME=z_{\alpha/2}\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ is the margin of error. Also, $ME=z_{\alpha/2}SE$, where $SE$ is the standard error.
Multiply $(1)$ by $-1$: $$-\bar{x}+ME>-\mu>-\bar{x}-ME \ \ \ \ \ (2)$$ Add $\mu+\bar{x}$ to $(2)$: $$\mu+ME>\bar{x}>\mu -ME \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ Hence, $(1),(2),(3)$ are equivalent.