### 4 Calculating other probabilities

In this Section we see how to calculate probabilities represented by areas other than those of the type shown in Figure 3.

#### 4.1 Case 1

Figure 4 illustrates what we do if both $Z$ values are positive. By using the properties of the standard normal distribution we can organise matters so that any required area is always of ‘standard form’.

**
Figure 4
**

##### Example 6

Find the probability that $Z$ takes values between 1 and 2.

##### Solution

Using Table 1:

$\text{P}\left(Z={z}_{2}\right)$ i.e. $\text{P}\left(Z=2\right)$ is 0.4772

$\text{P}\left(Z={z}_{1}\right)$ i.e. $\text{P}\left(Z=1\right)$ is 0.3413.

Hence $\text{P}\left(1<Z<2\right)=0.4772-0.3413=0.1359$

Remember that with a continuous distribution, $\text{P}\left(Z=1\right)$ is meaningless (will have zero probability) so that $\text{P}\left(1\le Z\le 2\right)$ is interpreted as $\text{P}\left(1<Z<2\right)$ .

#### 4.2 Case 2

The following diagram illustrates the procedure to be followed when finding probabilities of the form $\text{P}\left(Z>{z}_{1}\right)$ .

**
Figure 5
**

##### Example 7

What is the probability that $Z>2$ ?

##### Solution

$\text{P}\left(0<Z<2\right)=0.4772$ (from Table 1). Hence the probability is $0.5-0.4772=0.0228$ .

#### 4.3 Case 3

Here we consider the procedure to be followed when calculating probabilities of the form $\text{P}\left(Z<{z}_{1}\right)$ . Here the shaded area is the sum of the left-hand half of the total area and a ‘standard’ area.

**
Figure 6
**

##### Example 8

What is the probability that $Z<2$ ?

##### Solution

$\text{P}\left(Z<2\right)=0.5+0.4772=0.9772.$

#### 4.4 Case 4

Here we consider what needs to be done when calculating probabilities of the form

$\text{P}\left(-{z}_{1}<Z<0\right)$
where
${z}_{1}$
is positive. This time we make use of the symmetry in the standard normal distribution curve.

**
Figure 7
**

##### Example 9

What is the probability that $-2<Z<0$ ?

##### Solution

The area is equal to that corresponding to $\text{P}\left(0<Z<2\right)=0.4772.$

#### 4.5 Case 5

Finally we consider probabilities of the form $\text{P}\left(-{z}_{2}<Z<{z}_{1}\right)$ . Here we use the sum property and the symmetry property.

**
Figure 8
**

##### Example 10

What is the probability that $-1<Z<2$ ?

##### Solution

$$\begin{array}{rcll}\text{P}\left(-1<Z<0\right)=\text{P}\left(0<Z<1\right)& =& 0.3413& \text{}\\ \text{P}\left(0<Z<2\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}& =& 0.4772& \text{}\end{array}$$

Hence the required probability $\text{P}\left(-1<Z<2\right)$ is 0.8185.

Other cases can be handed by a combination of the ideas already used.

##### Task!

Find the following probabilities.

- $\text{P}\left(0<Z<1.5\right)$
- $\text{P}\left(Z>1.8\right)$
- $\text{P}\left(1.5<Z<1.8\right)$
- $\text{P}\left(Z<1.8\right)$
- $\text{P}\left(-1.5<Z<0\right)$
- $\text{P}\left(Z<-1.5\right)$
- $\text{P}\left(-1.8<Z<-1.5\right)$
- $\text{P}\left(-1.5<Z<1.8\right)$

(A simple sketch of the standard normal curve will help.)

- 0.4332 (direct from Table 1)
- $0.5-0.4641=0.0359$
- $\text{P}\left(0<Z<1.8\right)-\text{P}\left(0<Z<1.5\right)=0.4641-0.4332=0.0309$
- $0.5+0.4641=0.9641$
- $\text{P}\left(-1.5<Z<0\right)=\text{P}\left(0<Z<1.5\right)=0.4332$
- $\text{P}\left(Z<-1.5\right)=\text{P}\left(Z>1.5\right)=0.5-0.4332=0.0668$
- $\text{P}\left(-1.8<Z<-1.5\right)=\text{P}\left(1.5<Z<1.8\right)=0.0309$
- $\text{P}\left(0<Z<1.5\right)+\text{P}\left(0<Z<1.8\right)=0.8973$