The standard normal distribution and its table

When data is normally distributed, its standard scores are normally distributed too – with mean 0 and standard deviation 1, like the standard scores of any set of data. This distribution is called the standard normal distribution, and the standard score of a value chosen at random is written $\displaystyle Z$:

• The curve is symmetric round 0, and the area under it is 1.
• Every normal distribution turns into the standard one through the standard score, $\displaystyle {z}={\frac{x-\mu}{\sigma}}$: the part of the birth weights below 3,900 grams ($\displaystyle {\mu}={3300}$, $\displaystyle {\sigma}={500}$) equals the part of the standard scores below $\displaystyle {z}={\frac{3900-3300}{500}}={1.2}$.
• So one table, of the standard distribution, serves every normal distribution.

$\displaystyle \Phi(z)$ (read “phi of z”) is the area under the standard curve to the left of $\displaystyle z$ – the part of the standard scores less than $\displaystyle z$:

In the table below, $\displaystyle \Phi(1.2)$ is in row 1.2 and column 0: $\displaystyle {\Phi(1.2)}={0.8849}$. So 88.49% of babies are born weighing less than 3,900 grams.

Each cell of the table holds $\displaystyle \Phi(z)$, the area to the left of $\displaystyle z$. The table is cumulative: it gives the whole area up to $\displaystyle z$, not the area of one band.

• The row – $\displaystyle z$ to its first decimal place; the column – the second.
• $\displaystyle \Phi(0.47)$: row 0.4, column 7 – 0.6808.
• In a negative row the column goes further from 0: $\displaystyle \Phi(-1.23)$ is in row $\displaystyle -1.2$ and column 3 – 0.1093.
• Before reading the table, round $\displaystyle z$ to two decimal places.
• This table is on the course’s formula page too.

The whole area is 1, so the area to the right of $\displaystyle z$ is 1 minus the area to its left:

$\displaystyle {P(Z>1.2)}={1-0.8849}={0.1151}$.

By the symmetry, the area to the right of $\displaystyle z$ equals the area to the left of $\displaystyle -z$:

And indeed, in the table $\displaystyle {\Phi(-1.2)}={0.1151}$.

The area between $\displaystyle z_1$ and $\displaystyle z_2$ is the area to the left of $\displaystyle z_2$ minus the area to the left of $\displaystyle z_1$:

• $\displaystyle {P(-1<Z<2)}={\Phi(2)-\Phi(-1)}={0.9772-0.1587}={0.8185}$.
• Between $\displaystyle -1$ and 1: $\displaystyle {\Phi(1)-\Phi(-1)}={0.8413-0.1587}={0.6826}$ – about 68%, as in the 34/13.5/2.5 rule. The table gives the more exact number.
• $\displaystyle P(Z<z)$ and $\displaystyle P(Z\le z)$ are equal: a single point has no area.

$\displaystyle Z$ has the standard normal distribution. Find in the table and work out:

$\displaystyle P(Z<1.56)$

$\displaystyle P(Z>0.83)$

$\displaystyle P(Z<-0.5)$

$\displaystyle P(Z>-1.25)$

$\displaystyle P(-0.5<Z<1.5)$

Find in the table the standard score $\displaystyle z$ with 97.5% of the area to its left: $\displaystyle {\Phi(z)}={0.975}$.

Work out $\displaystyle P(-2<Z<2)$, and compare it with the 34/13.5/2.5 rule.