The Binomial and Poisson distributions were discrete distributions with the random variables taking on whole number or integer values.
We now look at a very important distribution which is continuous called the normal distribution. Here the probabilities found are between two values of the variable. This probability is represented by the area under the curve. This area can be found by integration but tables of values are available.
The histogram or bar graph of many frequency distributions is often shaped like a bell around the middle values.
As the width of each bar is made smaller and the size of the sample increases a smooth bellshaped curve results.
A bellshaped frequency distribution such as this is called a normal distribution.
A normal distribution and normal curve are symmetrical about the mode, median and mean.
Many distributions in real life are approximately normal distributions.
e.g. Heights and weights of a large group of people, lengths of leaves on a tree, number of peas in a pod etc.
The standard deviation gives a measure of the spread of the values around the mean.
The first normal curve below has a bigger standard deviation than the second.
The two parameters for the normal distribution are the mean,μ and the standard deviation, σ. To compare normal distributions both the mean and standard deviation must be known. The mean gives the centre and position of the curve and the standard deviation gives the shape. Nearly all of the values in a distribution are within three standard deviations of the mean.
Standard Normal Distribution
Because the shape of a normal distribution will vary according to the mean and standard deviation each distribution is changed to a standard normal curve.
Tables and calculators are available to find the probabilities associated with any position on a standard normal curve.
A standard normal curve has a mean of 0 and a standard deviation of 1.
The values from a distribution are standardised i.e. changed into standard scores or zscores using the formula:
where z is the number of standard deviations from the mean.
μ is the mean
x is the value being considered
σ is the standard deviation
This formula works out the number of standard deviations that a particular value is away from the mean. The probability associated with this zscore is then looked up in tables.
Finding Probabilities given a zscore
Below is a section of the Standard Normal Tables. For the full table click here.
Differences


Z

0

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

9

1.1

.3413

.3438

.3461

3485

.3508

.3531

.3554

.3577

.3599

.3621

2

5

7

9

12

14

16

18

21

1.1 
.3643

.3665

.3686

.3708

.3729

.3749

.3770

.3790

.3810

.3830

2

4

6

8

10

12

14

16

19

1.2

.3849

.3869

.3888

.3907

.3925

.3944

.3962

.3980

.3997

.4015

2

4

5

7

9

11

13

15

16

1.3

.4032

.4049

.4066

.4082

.4099

.4115

.4131

.4147

.4162

.4177

2

3

5

6

8

10

11

13

14

1.4

.4192

.4207

.4222

.4236

.4251

.4265

.4279

.4292

.4306

.4319

1

3

4

6

7

8

10

11

13

Example 1 Find the probability that a value chosen at random from a distribution lies in the shaded areas below.
The probability that a randomly chosen value lies in the shaded area is 0.3907 + 0.0013 = 0.3920 (shown on table above) i.e. P( 0 < z <1.237) = 0.3920 
The tables only cover positive values of z but the graph is symmetrical and probabilities are always positive.
Example 2 Find the probability that a value chosen at random from a distribution lies in the shaded area below.
Use the table above: P ( 1.3 < z < 1.4) = 0.4032 + 0. 4192 = 0.8224 
Finding Probabilities for "reallife" problems
For a reallife problem the values given have to be standardised and converted to zscores.Sketches of the normal distribution may help.
Example 1
The length of the life of a brand of light bulbs is normally distributed with an average life of 18 months and a standard deviation of 4 months.
a. Find the probability that a bulb chosen at random will have a life of more than 24 months
μ = 18 months
σ = 4 months
x = 24 months
On a diagram:
From tables:
shaded area, P(z > 1.5) = 0.5 − 0.4332 = 0.0668
The probability that a light bulb lives longer than 24 months is 0.0668
b. The number of months that you would expect 90% of the light bulbs to last for.
μ = 18 months
σ = 4 months
Step 1 Draw a diagram to show 90% of the normal curve
Step 2 The zscore of 1.281 above is worked out by finding 0.4 in the body of the tables.
Step 3 Now use the zscore formula
90% of light bulbs last more than 12.9 months.