How to calculate probability with Z score not on table?

Question

According to this Z table in my book, anything with a $z$ of over $3.16$ is probability 1, but this is not right. The textbook also has an example where $\mathbb P(Z\geq3.9) = 0.000048$; can somone explain to me how this answer was achieved?

Answer 1

Recall that a $z$-score is simply the number of standard deviations away from the mean ($\mu$). You said anything $>z=3.16$ will have a probability = 1. Indeed, this is not right. The probability is the area below the Normal distribution's curve.

For a score of $z=3.16$, the area under the Normal distribution from $-\infty \sigma$ to $3.16\sigma$ is $\approx 1$ (this is the probability). Note this an an estimate. There does exist a very small amount of area (again, synonymous with probability) above $3.16\sigma$. In your question, you state that $P(z\geq 3.9)= 0.000048$--this is that very small area ABOVE the $z$ score.

Consider the following picture, the $P(z<3.16)$ is shaded in and is easily detectable. I noted where the $P(z>3.16)$ should be, but could not shade it in because, well, it is quite small!

But how are $z$-scores calculated? As stated, it is the area below the Normal distribution's curve. This curve is defined by the probability density function $$\phi(x)=\frac{\exp(-0.5x^2)}{\sqrt{2\pi}} $$ (note this is for the special case of $\mu=0$ and $\sigma=1$). To find the probability for two $z$-scores $z_1$ and $z_2$ where $z_1<z_2$ we simply integrate between the two variables: $$\int_{z_1}^{z_2}\phi(x)dx. $$

Answer 2

For such values, you either need a table with more decimal places (as clearly 0.999952 was rounded off to 1), or more usually, a calculator that can compute probabilities for when $Z$ is large (sometimes referred to as tail probabilities).

As for how the answer was achieved, I'm not sure whether you mean how it came about from table (in which case it is impossible to calculate using the table you have), but mathematically:

$$\mathbb P [Z<z] = \int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{\frac{1}{2}z^2} dz$$

which you can only calculate by approximation, which is what the tables do for you (and subsequently round off).