Normal distribution, probability and modulus question

Question

Say $X$ is a random variable which is normally distributed with mean $0$ and variance $1$.

How do I find $k$ such that

$$\mathbb{P}(|X-k| < |X+k|) = 0.7$$

Answer 1

Such $k$ does not exist. We have that \begin{align*} |X+k|-|X-k| &=\frac{(|X+k|-|X-k|)(|X+k|+|X-k|)}{|X+k|+|X-k|}\\ &=\frac{|X+k|^2-|X-k|^2}{|X+k|+|X-k|}\\ &=\frac{4kX}{|X+k|+|X-k|}. \end{align*} Hence, $|X+k|-|X-k|>0$ if and only if $X>0$ when $k>0$ and $|X+k|-|X-k|>0$ if and only if $X<0$ when $k<0$. The probability does not depend on $k$ if $k\ne0$. We obtain $$ P(|X-k|<|X+k|)=\frac12 $$ if $k\ne0$ and it is equal to $0$ if $k=0$.

Answer 2

Hint:

If $k>0$ then:$$|x-k|<|x+k|\iff x>0$$

In words: to be more close to $k>0$ instead of $-k$ (on the same distance of $0$ as $k$, but on the other side) it is necessary and sufficient that $x>0$.

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edit:

Consequently $P(|X-k|<|X+k|)=P(X>0)=0.5\neq0.7$.

Conclusion: $k$ cannot be positive.

Likewise it can be shown that $k$ cannot be negative.

Evidently $P(|X|<|X|)=0\neq0.7$ so $k$ cannot take value $0$.

Final conclusion: no such $k$ exists.