Suppose $X$ is a random variable with finite mean and variance. For $50 \leq r<1$, we define the $r$ th percentile value $\pi_X(r)$ by
$$
P\left[X \leq \pi_X(r)\right]=\frac{r}{100} .
$$
Thus, the 90 th percentile value $\pi_X(90)$ is defined by
$$
P\left[X \leq \pi_X(90)\right]=0.90 .
$$
Show that
$$
\pi_X(90) \leq E[X]+3 \sigma,
$$
and
$$
\pi_X(95) \leq E[X]+\sigma \sqrt{19} .
$$