A level revision

Normal distribution

The normal distribution is a continuous probability distribution used to model continuous random variables - these variables that can take any value and are typically naturally occurring (e.g. height). As a continuous distribution the probability of, for example, a random person being 180cm tall is zero (i.e. P(X = 180) = 0). As the variable is continuous it is ranges you are concerned with such as 'What is the probability of a random person being smaller than 180cm?'

Shape of distribution

Normal distributions have a symmetrical bell shaped curve:

The area under the curve represents the probability with the total area being equal to 1.

Notation

If a continuous random variable X with a mean of and variance (where is the standard deviation) is normally distributed it is notated like so:

Example

The heights of men in a town can be modelled by the normal distribution with a mean of 182cm and a standard deviation of 10cm. This can be notated:

Standard normal distribution

When calculating the probability of a normal distribution it must be transformed from the standard normal distribution. The continuous random variable Z is used to denote the standard normal distribution. The standard normal distribution has a mean of 0 and a variance of 1:

In order to transform from the normal distribution variable X to the standard normal distribution variable Z the following equation is used:

Where is the mean, is the standard deviation (square root of the variance), is a particular value of the random variable X and is the corresponding value of random variable Z

Calculating probability

The probability is calculated using the following function:

This function gives the area under the curve (the probability) to the LEFT of the value z. That is to say it is cumulative, the value of is equal to the probability of the random variable Z being less than z. See examples for more information.

Values for this function for multiple values of z are found in the tables of values in your formula book.

Worked examples

The weight of adults in the UK is normally distributed with a mean of 15 Stone and a standard deviation of 3 stone. Find the probability that a randomly selected adult is:

1. Less than 18 stone
2. Less than 12 stone
3. Over 17 stone
4. Between 12 and 18 stone

1. Less than 18 stone.

First we need to transform the x value, 18, to the corresponding z value:

We want the probability that a random person is less than 18 stone which is given by the value of . From the formula tables this is 0.8413

2. Less than 12 stone

As before we must transform the x value to the corresponding z value:

The probability is given by the value but you will find that all values of z in your formula books are positive. As the graph is symmetrical the area under the curve to the left of -1 is the same as to the right of +1:

The values of give the value of the area to the left of the value z, in order to obtain the value to the right you simply subtract the value of from 1, the total area under the curve. As a result to get the value of any negative value of z:

So we simply plug our numbers in:

Using the value of from the formula book (or the previous question):

3. Over 17 stone

Firstly the value of z is calculated:

Because the question asked for the probability an adult weighs more than 17 stone the area under the graph needed is that to the right of the value of z (0.67). To obtain this value we simply subtract the value of , the value of the area to the left from 1, the total area:

4. Between 12 and 18 stone

The probability that a random adult weights between 12 and 18 stone is given by the area under the graph between the two corresponding values of z. From questions 1 and 3 they are 1 for 18 and -1 for 12. It always helps to draw a graph of the area you are looking for:

In this case subtracting the value of from will leave the area between the two values. From the previous questions the values are 0.1587 for and 0.8413 for . Therefore the probability is 0.8413 - 0.1587 = 0.6826.