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# Introduction and Tables

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## Introduction and Tables

**Continuous random variables** can take any values within a given range. Many naturally occurring phenomena can be approximated using a continuous random variable called the **Normal Distribution**.

* For example*: The heights of females or the weights of Russian ballet dancers.

In a normal distribution, much of the data is gathered around the mean. The distribution has a characteristic 'bell shape' symmetrical about the mean.

**Remember:** we write the symbol, μ, to stand for the **mean**.

The area of the bell shape = 1.

In any normal distribution, approximately 68% of the data will lie within one **standard deviation** of the mean. **The standard deviation** is an important measure of the spread of our data. The greater the standard deviation, the greater our spread of data.

**Remember:** we write the symbol, σ, to stand for the **standard deviation**.

**The standard deviation squared gives us the variance:**

Var(X) = σ^{2}

If X has a normal distribution with mean μ and variance σ^{2} then we write:

We will look closely at the normal distribution Z, with mean, μ = 0, and variance σ^{2} = 1.

Suppose for this distribution we wanted to calculate the P(Z < 1). Unlike with our discrete random variables we don't have a formula to work this out (there is one but it's way beyond the scope of A-Level maths!). The values of these probabilities have already been calculated and are tabulated in most statistics books. We will now look at how to read the tables to find these probabilities.

With Z~N(0,1) the P(Z ≤ z) = Φ(z)

Don't be put off with the Greek letter Φ (phi). Φ(z) just describes the area under the bell from that point!

**Example:**

For this example you will need a copy of the normal distribution tables.

By reading the tables we will calculate the probability of the following...

**a) (Z < 1.32)**

First we need to draw a bell.

To read Φ (1.32) we look down the first column until we reach 1.3 and then proceed across until we reach 1.32 (another part of the normal distribution tables).

Therefore, P(Z < 1.32) = Φ (1.32) = 0.9066

**b) P(Z > 1.32)**

Again, lets draw the bell.

This time, notice how we want the small area not the large. As the area is equal to 1 we subtract Φ (1.32) from 1 to obtain the probability.

So using result from before:

P(Z > 1.32) | = 1 − Φ (1.32) |

= 1 − 0.9066 | |

= 0.0934 |

**c) P(Z < −0.431)**

Hopefully you've guessed by now that the first step is to draw the bell!

You may have noticed that negative values are not given in the table. They are not needed as we can use the symmetry of the bell to calculate our answer.

Therefore, P(Z < -0.431) = 1 − Φ (0.431)

To read Φ (0.431) off the table we...

- Read down to 0.4.
- Read across to 0.43.
- Then look at 0.431 which says
**add 4**. We now add 4 to the last digit.

P(Z < -0.431) | = 1 − Φ (0.431) |

= 1 − 0.6668 | |

= 0.3332 |

You may not need an answer as accurate as this so it is often acceptable to round the z value to 2 d.p. In the case of the example above, reading Φ (0.43) would have been accepted as an approximate value to Φ (0.431).

**In the second example we met a general result:**

P(Z < -a) = 1 − Φ (a)

**Questions:**

1. Find the following probabilities:

Sometimes it is useful to find the value of z when given the probability. This is when we have to work backwards from the tables.

**Example:**

For this example you will need a copy of the normal distribution tables.

If Z~N(0,1) we are going to find the value of 'a', when:

- P(Z < a) = 0.7257
- P(Z > a) = 0.3594
- P(Z > a) = 0.9382

**Answers:**

Our first step will **always** be to draw the bell. Remembering that the distribution lies symmetrically about the mean, half of the area lies to the left and half lies to the right. This is vitally important with reading backwards from the tables as it helps us to place the value of 'a', determining if it's negative or positive.

1. P(Z < a) = 0.7257

To read backwards from the tables we simply look in the main 'belly' of the table. Looking to the left and upwards to obtain our solution.

2. P(Z > a) = 0.3594

3. P(Z > a) = 0.9382

**Try for yourself:**