Lecture 3Lecture 3

Chapter 2. Studying Normal Chapter 2. Studying Normal PopulationsPopulations

It is often the case that collected data have It is often the case that collected data have a distribution with the characteristic shape a distribution with the characteristic shape of the of the Normal distribution. Normal distribution.

Let’s have a look at an example…Let’s have a look at an example…

Example – Female HaematocritExample – Female Haematocrit

Haematocrit Haematocrit measures the percentage of measures the percentage of blood volume occupied by packed red blood volume occupied by packed red blood cells. blood cells.

Measurements taken from 126 female Measurements taken from 126 female medical students are as follows…medical students are as follows…

Female haematocrit measurementsFemale haematocrit measurements

42.0 42.0 40.0 45.0 42.0 43.0 36.0 43.0 40.0 42.0 35.042.0 42.0 40.0 45.0 42.0 43.0 36.0 43.0 40.0 42.0 35.0 46.0 45.0 38.0 40.0 46.0 45.0 43.0 42.0 41.0 44.0 41.046.0 45.0 38.0 40.0 46.0 45.0 43.0 42.0 41.0 44.0 41.0 49.0 42.0 35.0 46.0 44.0 37.0 43.0 42.0 41.0 36.0 41.049.0 42.0 35.0 46.0 44.0 37.0 43.0 42.0 41.0 36.0 41.0 44.0 38.0 40.5 42.0 38.0 35.0 42.0 42.0 40.0 45.0 40.044.0 38.0 40.5 42.0 38.0 35.0 42.0 42.0 40.0 45.0 40.0 55.0 45.0 44.0 44.0 43.0 42.0 42.0 43.0 42.0 45.0 43.055.0 45.0 44.0 44.0 43.0 42.0 42.0 43.0 42.0 45.0 43.0 44.0 41.0 46.0 49.0 44.0 44.0 32.5 42.0 44.0 49.0 41.544.0 41.0 46.0 49.0 44.0 44.0 32.5 42.0 44.0 49.0 41.5 48.0 42.0 39.0 45.0 41.0 42.0 43.0 46.0 40.0 43.0 42.048.0 42.0 39.0 45.0 41.0 42.0 43.0 46.0 40.0 43.0 42.0 44.0 40.0 44.0 41.0 44.0 43.0 49.0 40.5 39.0 48.0 40.044.0 40.0 44.0 41.0 44.0 43.0 49.0 40.5 39.0 48.0 40.0 41.0 41.0 41.0 45.0 36.0 39.0 36.0 38.0 45.0 46.0 41.041.0 41.0 41.0 45.0 36.0 39.0 36.0 38.0 45.0 46.0 41.0 40.0 43.0 34.5 42.0 42.0 39.0 41.5 46.0 42.0 44.0 46.040.0 43.0 34.5 42.0 42.0 39.0 41.5 46.0 42.0 44.0 46.0 44.0 40.0 38.5 40.0 40.0 44.0 39.0 40.0 39.0 43.0 42.044.0 40.0 38.5 40.0 40.0 44.0 39.0 40.0 39.0 43.0 42.0 36.0 46.0 44.5 48.0 45.536.0 46.0 44.5 48.0 45.5

Let’s look at the shape of the distribution of this data Let’s look at the shape of the distribution of this data using a histogram…using a histogram…

Example – Female HaematocritExample – Female Haematocrit

Haematocrit (F)

Frequency

56524844403632

35

30

25

20

15

10

5

0

Histogram of Haematocrit (F)

These data show the characteristic shape of the These data show the characteristic shape of the Normal Distribution.Normal Distribution.

It is characterised by the symmetrical “bell It is characterised by the symmetrical “bell shape”, which corresponds to values near the shape”, which corresponds to values near the mean being more common, while values further mean being more common, while values further away “tail off” in terms of their frequencies.away “tail off” in terms of their frequencies.

A perfect Normal distribution curve looks like….A perfect Normal distribution curve looks like….

Standard Deviations from Mean

Frequency

Curv

e

43210-1-2-3-4

0.4

0.3

0.2

0.1

0.0

Normal Distribution Frequency Curve

In order to understand what it really means for In order to understand what it really means for data to be Normally distributed, we first need to data to be Normally distributed, we first need to consider the idea of consider the idea of probabilityprobability……

ProbabilityProbability ProbabilityProbability is used to measure the likelihood of is used to measure the likelihood of

an event occurring.an event occurring.

DefinitionDefinitionSuppose we were to repeat a particular experiment over Suppose we were to repeat a particular experiment over and over again. and over again.

Then the Then the probabilityprobability of a particular outcome of a particular outcome AA is is defined as the defined as the proportionproportion of the total number of repeats of the total number of repeats in which in which AA would actually occur, if we were to keep on would actually occur, if we were to keep on repeating the experiment. repeating the experiment.

We denote this probability by We denote this probability by Pr(A)Pr(A). .

Probability ExamplesProbability Examples1.1. Rolling a fair dieRolling a fair die

We roll a standard six-sided die. Let event A be that We roll a standard six-sided die. Let event A be that the die lands with three spots face up. the die lands with three spots face up.

Then the probability of the event A is:Then the probability of the event A is:

Pr(A) = 1/6 ≈ 0.167Pr(A) = 1/6 ≈ 0.167

because in the long run, the proportion of times that A because in the long run, the proportion of times that A happens will be 1/6.happens will be 1/6.

Note that in this experiment there are six equally likely Note that in this experiment there are six equally likely outcomes, all with probability 1/6. outcomes, all with probability 1/6.

Probability ExamplesProbability Examples2.2. Tossing a fair coinTossing a fair coin

You toss a fair coin once. Let event A be that the coin lands You toss a fair coin once. Let event A be that the coin lands heads up. heads up.

Then the probability of the event A is:Then the probability of the event A is:

Pr(A) = ½ = 0.5Pr(A) = ½ = 0.5

because in the long run, the proportion of times that A because in the long run, the proportion of times that A happens will be 1/2. This time there are two possible happens will be 1/2. This time there are two possible outcomes with equal probability. outcomes with equal probability.

Note that the Probability scale runs between 0 and 1 Note that the Probability scale runs between 0 and 1 inclusive. The higher the number, the more likely the inclusive. The higher the number, the more likely the event. event.

Probability ExamplesProbability Examples3.3. Buying a ticket for the UK National LottoBuying a ticket for the UK National Lotto

You buy a single ticket for one draw of the UK National You buy a single ticket for one draw of the UK National Lotto. The event A is that your six numbers exactly Lotto. The event A is that your six numbers exactly match the six main numbers drawn from 1, … , 49, so match the six main numbers drawn from 1, … , 49, so that you win a share of the jackpot. that you win a share of the jackpot.

Then the probability of the event A is:Then the probability of the event A is:

Pr(A) = 1 / 13,983,816 ≈ 0.0000000715Pr(A) = 1 / 13,983,816 ≈ 0.0000000715

because there are 13,983,816 equally likely outcomes because there are 13,983,816 equally likely outcomes for the six main numbers. for the six main numbers.

Probability measurements only really make Probability measurements only really make sense for sense for discretediscrete outcomes, i.e. when we can outcomes, i.e. when we can make a make a listlist of of all the possible outcomes. all the possible outcomes.

When the measurements are on a continuous When the measurements are on a continuous scale, such as the haematocrit measures, then scale, such as the haematocrit measures, then there are infinitely many possible outcomes, and there are infinitely many possible outcomes, and it is not possible to list them. it is not possible to list them.

The distribution of haematocrit outcomes has The distribution of haematocrit outcomes has roughly the Normal distribution shape: roughly the Normal distribution shape:

Haematocrit (%)

Pro

babili

ty D

ensi

ty

555045403530

0.12

0.10

0.08

0.06

0.04

0.02

0.00

Fitted Normal Distribution for Female Haematocrit