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 freuqency 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 or the spread of the values around the mean.
The first normal curve below has a bigger standard deviation than the second.
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.
Probabilities and Normal Distributions
Data associated with a normal distribution has the following features.
Approximately 68% of the data lies within ONE standard deviation of the mean.

Approximately 95% of the data lies within TWO standard deviations of the mean.

Approximately 99% of the data lies within THREE standard deviations of the mean.

This means that, for example, a value picked at random from a set of data would have a probability of 0.95 of being within 2 standard deviations of the mean. 
Example
The times for a bus journey recorded over a year are normally distributed with a mean of 35 minutes and a standard deviation of 9 minutes. What is the probability that a bus selected at random takes a. between 26 minutes and 44 minutes? b. Between 17 and 62 minutes? 
The mean μ is 35 and the standard deviation σ is 9 a. 26 is μ − σ and 44 is μ + σ Therefore there is a probability of 0.68 that the bus takes between 26 and 44 minutes. b. 17 is μ − 2σ which is equivalent to a probability of 0.475 and 62 is μ + 3σ which is equivalent to a probability of 0.495. Therefore there is a probability of 0.475 + 0.495 = 0.97 that a journey takes between 17 and 62 minutes. 
Standard Normal Distribution
Tables and calculators are available to find the probabilities associated with any position on astandard normal curve. A standard normal curve has a mean of 0 and a standard deviation of 1. The values from a distribution are changed to 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 normallly 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.