For $z >0$, the right tail of the standard normal distribution (that is, the area to the right of $z$) which is often denoted by $Q(z)$ is bounded as follows:

$$\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z} - \frac{1}{z^3}\right ) \ < \ Q(z) \ < \ \frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right ).$$

See, for example this answer on math.SE for a proof. The bounds blow up/down to $\pm \infty$ as $z \to 0$ but are quite useful in the regions not covered by typical tables of the cumulative distribution function of the standard normal random variable. For example, $$0.0000873 < Q(3.75) < 0.0000940$$ while the actual value is slightly larger than $0.0000884$

For $z >0$, the right tail of the standard normal distribution (that is, the area to the right of $z$) which is often denoted by $Q(z)$ is bounded as follows:

$$\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z} - \frac{1}{z^3}\right ) \ < \ Q(z) \ < \ \frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right ).$$

See, for example this answer on math.SE for a proof. The bounds blow up/down to $\pm \infty$ as $z \to 0$ but are quite useful in the regions not covered by typical tables of the cumulative distribution function of the standard normal random variable. For example, $$0.0000873 < Q(3.75) < 0.0000940$$ while the actual value is slightly larger than $0.0000884$

For $z >0$, the right tail of the standard normal distribution (that is, the area to the right of $z$) which is often denoted by $Q(z)$ is bounded as follows:

$$\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z} - \frac{1}{z^3}\right ) \ < \ Q(z) \ < \ \frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right ).$$

See, for example this answer on math.SE for a proof. The bounds blow up/down to $\pm \infty$ as $z \to 0$ but are quite useful in the regions not covered by typical tables of the cumulative distribution function of the standard normal random variable.

For $z >0$, the right tail of the standard normal distribution (that is, the area to the right of $z$) which is often denoted by $Q(z)$ is bounded as follows:

$$\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z} - \frac{1}{z^3}\right ) \ < \ Q(z) \ < \ \frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right ).$$

See, for example this answer on math.SE for a proof. The bounds blow up/down to $\pm \infty$ as $z \to 0$ but are quite useful in the regions not covered by typical tables of the cumulative distribution function of the standard normal random variable. For example, $$0.0000873 < Q(3.75) < 0.0000940$$ while the actual value is slightly larger than $0.0000884$