Normal distribution
1.
Learn about the properties of a normal
distribution
2.
Solve problems using tables of the normal
distribution
3.
Meet some other examples of continuous
probability distributions
Types of variables
Discrete & Continuous
Describe what types of data could be
described as continuous random
variable X=x
– Arm lengths
– Eye heights
REMEMBER your box plot
Middle 50%
LQ
UQ
range
The Normal Distribution
(Bell Curve)
Average contents 50
Mean = μ = 50
Standard deviation = σ = 5
The normal distribution is a
theoretical probability
the area under the curve adds up to one
The normal distribution is a
A Normal distribution is a theoretical model of the
wholetheoretical
population. It isprobability
perfectly symmetrical about
the central value; the mean μ represented by zero.
the area under the curve adds up to one
The X axis is divided up into deviations from the
As well
as the
theshaded
meanarea
theisstandard
mean.
Below
one deviation
deviation
(σ) must also be known.
from
the mean.
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Two standard deviations
from the mean
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Three standard deviations
from the mean
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
A handy estimate – known as the
Imperial Rule for a set of normal data:
68% of data will fall within 1σ of the μ
P( -1
<
z
<
1
)
=
0.683
=
0
1
2
68.3%
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
3
4
5
95% of data fits within 2σ of the μ
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
P( -2 <
-3
-2
z
<
-1
2
0
)
1
=
2
0.954
3
=
4
95.4%
5
99.7% of data fits within 3σ of the μ
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
P( -3 < z < 3
-1
0
)
=
1
2
0.997
3
=
4
5
99.7%
Simple problems solved using the imperial
rule - firstly, make a table out of the rule
<-3
0%
-3 to - -2 to 2
1
2%
14%
-1 to
0
0 to 1
1 to 2
2 to 3
>3
34%
34%
14%
2%
0%
The heights of students at a
college were found to follow
a bell-shaped distribution
with μ of 165cm and σ of 8
cm.
What proportion of students
are smaller than 157 cm
first standardis e
x

16%
z
157  165
first 157cm is
 1
8
or 1 below the 
Simple problems solved using the imperial
rule - firstly, make a table out of the rule
<-3
0%
-3 to - -2 to 2
1
2%
14%
-1 to
0
0 to 1
1 to 2
2 to 3
>3
34%
34%
14%
2%
0%
The heights of students at a
college were found to follow
a bell-shaped distribution
with μ of 165cm and σ of 8
cm.
Above roughly what height
are the tallest 2% of the
students?
The tallest 2% of students are beyond 2 of 
165 + 2 x 8 = 181 cm
Task – class 10 minutes
finish for homework

Exercise A
Page 76
The Bell shape curve happens
so when recording continuous
random variables that an
equation is used to model the
shape exactly.
1  12 x2
y
e
2
Put it into your
calculator and use the
graph function.
Sometimes you will see
it using phi =.
1  12 z 2
 ( z) 
e
2
Luckily you don’t have to use the equation each time and you
don’t have to integrate it every time you need to work out the
area under the curve – the normal distribution probability
There are normal distribution tables
How to read the Normal distribution
table
Φ(z) means the area under the curve
on the left of z
How to read the Normal distribution
table
Φ(0.24) means the area under the
curve on the left of 0.24 and is this
value here:
Values of Φ(z)

Φ(-1.5)=1- Φ(1.5)
Values of Φ(z)


Φ(0.8)=0.78814 (this is for the left)
Area = 1-0.78814 = 0.21186
Values of Φ(z)


Φ(1.5)=0.93319
Φ(-1.00)
=1- Φ(1.00)
=1-0.84134
=0.15866

Shaded area =
Φ(1.5)- Φ(-1.00)
= 0.93319 - 0.15866
= 0.77453
Task

Exercise B page 79
Solving Problems using
the tables
NORMAL DISTRIBUTION
 The area under the curve is the probability of getting less than the z
score. The total area is 1.
 The tables give the probability for z-scores in the distribution
X~N(0,1), that is mean =0, s.d. = 1.
ALWAYS SKETCH A DIAGRAM
 Read the question carefully and shade the area you want to find. If
the shaded area is more than half then you can read the probability
directly from the table, if it is less than half, then you need to
subtract it from 1.
NB If your z-score is negative then you would look up the positive from
the table. The rule for the shaded area is the same as above: more
than half – read from the table, less than half subtract the reading
from 1.
You will have to standardise if the
mean is not zero and the standard
deviation is not one
Task

Exercise C page 168
Normal distribution
problems in reverse


Percentage points table on page 155
Work through examples on page 84
and do questions Exercise D on page
85
Key chapter points



The probability distribution of a continuous random
variable is represented by a curve. The area under
the curve in a given interval gives the probability of
the value lying in that interval.
If a variable X follows a normal probability
distribution, with mean μ and standard deviation σ,
we write X ̴ N (μ, σ2)
The variable Z=
is called the standard
normal variable corresponding to X
Key chapter points cont.


If Z is a continuous random variable
such that Z ̴ N (0, 1) then
Φ(z)=P(Z<z)
The percentage points table shows, for
probability p, the value of z such that
P(Z<z)=p