# Given that z is a standard normal random variable, compute the following probabilities

– $P (z \space \leq \space – \space 1.0 )$

– $P (z \space \geq \space – \space 1 )$

– $P (z \space \geq \space – \space 1.5 )$

– $P ( – \space 2.5 \space \geq \space \space z )$

– $P (- \space 3 \space < \space z \space \geq \space \space 0 )$

The main objective of this question is to find the probabilities for the given expressions given the z score, which is a standard random variable.

Single constant number

Random number

This question uses the concept of z-score. The standard normal z-table is the abbreviation for the z-table. Standard Normal models are used in hypothesis testing as well as the differences between two means.  $100 \space %$ of an area under a distribution of normal curve is represented by a value of one hundred percent or $1$. The z-table tells us how much of the curve is below a given point. The z-score is calculated as:

$\space z \space = \frac{ score \space – \space mean }{ standard deviation}$

Probability

We have to compute the probabilities.

a) From the z-table, we know that the value of $– \space 1$ is:

$\space = \space 0.1587$

So:

$\space P (z \space \leq \space – \space 1.0 ) \space = \space 0.1587$

b) Given that:

$\space P (z \space \geq \space – \space 1 )$

Thus:

$\space = \space 1 \space – \space P (z \space \leq \space – \space 1 )$

We know that:

$\space P (z \space \leq \space – \space 1.0 ) \space = \space 0.1587$

So:

$\space = \space 1 \space – \space 0.1587$

$\space = \space 0.8413$

c) Given that:

$\space P (z \space \geq \space – \space 1.5 )$

So:

$\space = \space 1 \space – \space P(z \space \leq \space – \space 1.5$

$\space = \space 1 \space – \space 0.0668$

$\space = \space 0.9332$

d) Given that:

$\space P ( – \space 2.5 \space \geq \space \space z )$

So:

$\space P(z \space \geq \space – \space 2.5)$

$\space 1 \space – \space P(z \space \leq \space – \space 2.5)$

$\space = \space 1 \space – \space 0.0062$

$\space = \space 0.9938$

e) Given that:

$\space P (- \space 3 \space < \space z \space \geq \space \space 0 )$

So:

$\space P(z \space \leq \space 0) \space – \space P(z \leq \space – \space 3)$

$\space 0.5000 \space – \space 0.0013$

$\space = \space 0.4987$

The probability for the $P (z \space \leq \space – \space 1.0 )$  is:

$\space = \space 0.1587$

The probability for the $P (z \space \geq \space – \space 1 )$  is:

$\space = \space 0.8413$

The probability for the $P (z \space \geq \space – \space 1.5 )$  is:

$\space = \space 0.9332$

The probability for the $P ( – \space 2.5 \space \geq \space \space z )$  is:

$\space = \space 0.9938$

The probability for the $P (- \space 3 \space < \space z \space \geq \space \space 0 )$  is:

$\space = \space 0.4987$

## Example

Find the probability for  $z$ which is a standard random variable.

$\space P (z \space \leq \space – \space 2.0 )$

We have to compute the probabilities. From the z-table, we know that the value of $– \space 2$ is:

$\space = \space 0.228$

So:

$\space P (z \space \leq \space – \space 1.0 ) \space = \space 0.228$