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“Teach A Level Maths”
Statistics 1
The Normal Distribution
© Christine Crisp
The Normal Distribution
Statistics 1
AQA
Edexcel
Normal Distribution diagrams in the examples and exercises in this presentation have
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The Normal Distribution
Suppose we have a crate of apples which are to be sorted
by weight into small, medium and large. If we wanted 25%
to be in the large category, we would need to know the
lowest weight a “large” apple could be.
( Here I am using weight in the everyday sense; the
quantity measured in kilograms and grams. If you are a
physicist you will refer to mass. )
To solve a problem like this we can use a statistical model.
A model often used for continuous quantities such as
weight, volume, length and time is the Normal Distribution.
The Normal Distribution is an example of a probability
model.
The Normal Distribution
Characteristics of the Normal Distribution
If we were to show the weights of a large number of our
apples in a histogram we might get this:
There are not many very light . . .
The distribution is fairly symmetric.
or very heavy apples.
The Normal Distribution
Characteristics of the Normal Distribution
The Normal distribution model is a symmetric bell-shaped
curve. We fit it as closely as possible to the data.
The Normal
Distribution curve
To fit the curve we use the mean, m, and variance,
the data. These are the parameters of the model.
 2, of
If X is the random variable “ the weight of apples”, we
write
X ~ N (m ,  2 )
Reminder:  is the standard deviation.
The Normal Distribution
Characteristics of the Normal Distribution
The Normal distribution model is a symmetric bell-shaped
curve. We fit it as closely as possible to the data.
The Normal
Distribution curve
For this curve we might have
If a question gives me the
2
X ~ N ( 350, 110 )
standard deviation, I often
write the variance in this
form instead of simplifying.
The Normal Distribution
The axis of symmetry of the Normal distribution passes
through the mean.
e.g.
X ~ N (4, 1)
X ~ N (5, 1)
The Normal Distribution
A smaller variance “squashes” the distribution closer to
the mean.
e.g.
X ~ N (4, 0  5 2 )
X ~ N (4, 1)
The Normal Distribution
Finding probabilities
When we had a discrete distribution we could find a
probability by using a formula.
e.g. The r.v. X has probability distribution function
(p.d.f.) given by
x
P( X  x) 
6
x
1
P (X = x)
1
6
2
2
6
for
3
x  1, 2, 3
05
P( X  x)
3
6
x
1
2
3
For a continuous distribution, a probability is given by an
area under the graph of the p.d.f.
The Normal Distribution
For example, the probability that an apple taken at
random weighs less than 200 grams is given by the area
to the left of a line through 200.
X ~ N ( 350, 1102 )
P( X  200)
The total area gives the sum of the probabilities so
equals 1.
 ( x  m )2
The p.d.f. of the Normal curve is
f ( x) 
1
 2
e
2 2
Since this is a difficult function the probabilities have been
worked out and listed in a table.
The Normal Distribution
Finding Probabilities
Before we do an example, find the table of probabilities
in your formulae book.
The table gives probabilities for the random variable Z
where
Z ~ N (0, 1)
Notice the diagram at the top of the page. The shading
shows that the probabilities given are always less than
the z value used.
( I will usually write the values to 4 d.p. )
Since we may need to use these values to find others, we
will always draw a sketch.
The Normal Distribution
e.g.1 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  6) (c) P (1  Z  2)
Solution: (a)
P ( Z  1)
Z ~ N (0, 1)
(use the table )
 0  8413
0 1
(b) P ( Z  1  6)
 1  P ( Z  1  6)
P ( Z  1  6) is a lot to write so we
can write (1·6) which means the
area to the left of 1·6.
Z ~ N (0, 1)
(1  6)
0 1 6
The Normal Distribution
e.g.1 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  6) (c) P (1  Z  2)
Solution: (a)
P ( Z  1)
Z ~ N (0, 1)
(use the table )
 0  8413
0 1
(b) P ( Z  1  6)
 1  (1  6)
 1  0  9452
 0  0548
Z ~ N (0, 1)
(1  6)
Tip: It’s useful to always check
0 1 6
the answer: an area less than 0·5
corresponds to less than half the area and vice versa.
The Normal Distribution
e.g.1 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  6) (c) P (1  Z  2)
Z ~ N (0, 1)
(c)
P (1  Z  2)
Solution:
 ( 2)  (1)
0 1 2
Z ~ N (0,1)
( 2 )
0 1 2
The Normal Distribution
e.g.1 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  6) (c) P (1  Z  2)
Z ~ N (0, 1)
(c)
P (1  Z  2)
Solution:
 ( 2)  (1)
(use the table )
 0  9773  0  8413
0 1 2
Z ~ N (0,1)
(1)
 0  1360
0 1 2
The Normal Distribution
Special Cases
•
The probability of any single value is zero.
e.g.
P ( Z  1)  0
Z ~ N (0, 1)
( There is no area. )
0 1
•
Also, using
e.g.
 is the same as using <
P ( Z  1)  P ( Z  1)
Neither property holds for discrete distributions.
The Normal Distribution
SUMMARY
 The Normal Distribution is continuous and symmetric
about the mean.
 An area under the curve gives a probability.
 The formula booklet gives a table of probabilities
for the random variable Z where
Z ~ N (0, 1)
 The table gives values of P ( Z  z ). This is written
as (z ).
P( Z  z)
(z )
0
z
 When using the table we always draw a sketch
showing the required probability.
The Normal Distribution
e.g. 2 Find the percentage of the Normal distribution
that lies within 1 standard deviation on either
side of the mean.
Solution: Since we want   1 ( the standard deviation ),
the variance,  ,2 is also equal to 1.
The mean can be any value, so let Z ~ N (0, 1) .
We can find the percentage from the probability, so we
need the probability that Z is between – 1 and + 1.
Z
Z
P ( 1  Z  1)
1
0
1
1
The easiest way to find this area is to find
and subtract ( 0) . . .
0
1
(1) .
. .
The Normal Distribution
e.g. 2 Find the percentage of the Normal distribution
that lies within 1 standard deviation on either
side of the mean.
Solution: Since we want   1 ( the standard deviation ),
the variance,  ,2 is also equal to 1.
The mean can be any value, so let Z ~ N (0, 1) .
We can find the percentage from the probability, so we
need the probability that Z is between – 1 and + 1.
Z
Z
P ( 1  Z  1)
1
0
1
1
The easiest way to find this area is to find
and subtract ( 0) . . . then multiply by 2.
0
1
(1) .
. .
The Normal Distribution
e.g. 2 Find the percentage of the Normal distribution
that lies within 1 standard deviation on either
side of the mean.
Solution: Since we want   1 ( the standard deviation ),
the variance,  ,2 is also equal to 1.
The mean can be any value, so let Z ~ N (0, 1) .
We can find the percentage from the probability, so we
need the probability that Z is between – 1 and + 1.
Z
Z
P ( 1  Z  1)
1
0
Can you see what
1
(0) equals without
(0)  0  5
1
0
1
using the table?
The Normal Distribution
e.g. 2 Find the percentage of the Normal distribution
that lies within 1 standard deviation on either
side of the mean.
Solution: Since we want   1 ( the standard deviation ),
the variance,  ,2 is also equal to 1.
The mean can be any value, so let Z ~ N (0, 1) .
We can find the percentage from the probability, so we
need the probability that Z is between – 1 and + 1.
Z
Z
P ( 1  Z  1)
1
0
1
1
0
1
(1)  (0)  0  8413  0  5000  0  3413
P (1  Z  1)  2  0  3413  0  6826
The Normal Distribution
e.g. 2 Find the percentage of the Normal distribution
that lies within 1 standard deviation on either
side of the mean.
Solution: Since we want   1 ( the standard deviation ),
the variance,  ,2 is also equal to 1.
The mean can be any value, so let Z ~ N (0, 1) .
We can find the percentage from the probability, so we
need the probability that Z is between – 1 and + 1.
Z
P ( 1  Z  1)
1
P (1  Z  1)  0  6826
0
1

The percentage is
approximately 68%.
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1  3) (b) P ( Z  1  45) (c) P (0  8  Z  1  8)
2. Find the percentage of the Normal distribution that
lies within (a) 2 standard deviations either side of
the mean and (b) 3 standard deviations either side
of the mean.
The Normal Distribution
Solutions:
1(a)
P ( Z  1  3)
 (1  3)
 0  9032
Z
0 1 3
1(b)
P ( Z  1  45)
Z
 1  (1  45)
 1  0  9265
 0  0735
0
1  45
The Normal Distribution
Solutions:
1(c)
P (0  8  Z  1  8)
Z
 (1  8)  (0  8)
 0  9641  0  7881
 0  1760
0 0 8 1 8
The Normal Distribution
2. Find the percentage of the Normal distribution that
lies within (a) 2 standard deviations either side of
the mean and (b) 3 standard deviations either side
of the mean.
Solution: Let
Z ~ N (0, 1)
(a) We want
P (2  Z  2)
( 2)  (0)  0  97725  0  5000
 0  47725
 P (2  Z  2)  2  0  47725
 0  9545
Z
2
0
2
Approximately 95% of the Normal distribution
lies within 2 standard deviations of the mean.
(b) The method is the same. The answer is approx. 99·8%.
The Normal Distribution
SUMMARY
 The percentages of the Normal Distribution lying within
the given number of standard deviations either side of
the mean are approximately:
1 s.d. : 68%
68%
 m
2 s.d. : 95%
3 s.d. : 99·8%
99·8%
95%
 2

m
2
 3
m
3
The Normal Distribution
e.g.3 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  5)
(c)
P (1  Z  2)
Solution:
(a) P ( Z  1)
Z
1 0
The table only gives
probabilities for positive z
values so we have to find
an equal area that is in
the table.
P ( Z  1)  P ( Z  1)
 (1)
 0  8413
Z
1 0 1
The Normal Distribution
Solution:
(b) P ( Z  1  5)
Z
 1 5 0
This area equals
 1  (1  5)
 1  0  9332
 0  0668
Z
0 1 5
The Normal Distribution
Solution:
(c) P ( 1 
Z  2)
Z
 ( 2)  (1)
Tip: Work out (1)
first or you could
make a sign error.
( 1)    (1)
 1  0  8413
 0  1587
So,
1 0
2
Z
  (1)
(1)
1 0 1
(2)  ( 1)  0  9773  0  1587
 0  8186
The Normal Distribution
SUMMARY
 To find (  z ) , the area to the left of a negative
number, we use
(  z )  1  ( z )
N.B. The procedure for this and all other areas involving
negative values can be seen from the diagram.
NEVER try to do these questions without at least
one diagram.
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
find (a)
(c)
Z ~ N (0, 1)
(b) P ( Z  1)
P ( Z  1  3)
P (2  Z  1  2) (d) P (1  26  Z  0  34)
There are 2 methods of doing part (d). See if
you can spot them both and use the quicker.
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
find (a)
(c)
Z ~ N (0, 1)
(b) P ( Z  1)
P ( Z  1  3)
P (2  Z  1  2) (d) P (1  26  Z  0  34)
Solution:
(a) P ( Z  1  3)
Z
Z
=
 1 3
0
0
P ( Z  1  3)  (1  3)
 0  9032
1 3
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
find (a)
(c)
Solution:
(b) P ( Z
Z ~ N (0, 1)
(b) P ( Z  1)
P ( Z  1  3)
P (2  Z  1  2) (d) P (1  26  Z  0  34)
 1)
Z
Z
=
1
0
P ( Z  1)  1  (1)  1  0  8413
 0  1587
0
1
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
find (a)
(c)
Z ~ N (0, 1)
(b) P ( Z  1)
P ( Z  1  3)
P (2  Z  1  2) (d) P (1  26  Z  0  34)
Solution: (c)
P (2  Z  1  2)
(1  2)
Z
0
2
0
2)
1 2
P (2  Z  1  2)  (1  2)  ( 2)
( 2)  1  ( 2)  1  0  9773  0  0227
P(2  Z  1  2)  0  8849  0  0227
 0  8622
2
0
1 2
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
find (a)
(c)
Z ~ N (0, 1)
(b) P ( Z  1)
P ( Z  1  3)
P (2  Z  1  2) (d) P (1  26  Z  0  34)
Solution: (d)
P (1  26  Z  0  34)
Method 1:
Area equals (1  26)  (0  34)
Z
 1  26
 0  34
0  34
1  26
 0  8962  0  6331
 0  2631
The Normal Distribution
Exercise
1. If Z is a random variable with distribution
find (a)
(c)
(b) P ( Z  1)
P ( Z  1  3)
P (2  Z  1  2) (d) P (1  26  Z  0  34)
Solution: (d)
Z
 1  26
 0  34

Z ~ N (0, 1)
Method 2:
P (1  26  Z  0  34)
P (1  26  Z  0  34) 
( 0  34)  ( 1  26)
( 0  34)  1  (0  34)
 1  0  6331
 0  3669
( 1  26)  1  (1  26)
 1  0  8962
 0  1038
P(1  26  Z  0  34)  0  3669  0  1038  0  2631
The Normal Distribution
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
The Normal Distribution
SUMMARY
 The Normal Distribution is continuous and symmetric
about the mean.
 An area under the curve gives a probability.
 The formula booklet gives a table of probabilities
for the random variable Z where
Z ~ N (0, 1)
 The table gives values of P ( Z  z ). This is written
as (z ).
P( Z  z)
(z )
0
z
 When using the table we always draw a sketch
showing the required probability.
The Normal Distribution
Special Cases
•
The probability of any single value is zero.
e.g.
P ( Z  1)  0
Z
( There is no area )
0 1
•
Also, using
e.g.
 is the same as using <
P ( Z  1)  P ( Z  1)
Neither property holds for discrete distributions.
The Normal Distribution
e.g.1 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  6) (c) P (1  Z  2)
Solution: (a)
Z
P ( Z  1)
(use the table )
 0  8413
0 1
(b) P ( Z  1  6)  1  (1  6)
Z
P ( Z  1  6)
Z
(1  6)
0 1 6
 1  0  9452  0  0548
0 1 6
The Normal Distribution
Solution:
(c)
P (1  Z  2)
Z
 ( 2)  (1)
P (1  Z  2)
0 1 2
Z~
( 2 )
0 1 2
Z
(1)
0 1 2
 0  9773  0  8413  0  1360
The Normal Distribution
e.g. 2 Find the percentage of the Normal distribution
that lies within 1 standard deviation on either
side of the mean.
Solution: Let Z ~ N (0, 1)
We can find the percentage from the probability, so we
need the probability that Z is between m – 1 and m + 1
where m is the mean and equals zero.
Z
Z
P ( 1  Z  1)
1
0
1
1
0
1
(1)  (0)  0  8413  0  5000  0  3413
P (1  Z  1)  2  0  3413  0  6826
 The percentage is approximately 68%.
The Normal Distribution
SUMMARY
 The percentages of the Normal Distribution lying within
the given number of standard deviations either side of
the mean are approximately:
1 s.d. : 68%
68%
 m
2 s.d. : 95%
3 s.d. : 99·8%
99·8%
95%
 2

m
2
 3
m
3
The Normal Distribution
e.g.3 If Z is a random variable with distribution
Z ~ N (0, 1)
find (a) P ( Z  1) (b) P ( Z  1  5)
(c)
P (1  Z  2)
Solution:
(a) P ( Z  1)
Z
1 0
The table only gives
probabilities for positive z
values so we have to find
an area equal to this that
is in the table.
P ( Z  1)  P ( Z  1)
 (1)
 0  8413
Z
1 0 1
The Normal Distribution
Solution:
(b) P ( Z  1  5)
Z
 1 5 0
This area equals
 1  (1  5)
 1  0  9332
 0  0668
Z
0 1 5
The Normal Distribution
Solution:
(c) P ( 1 
Z  2)
Z
 ( 2)  (1)
Tip: Work out
first.
(1)
( 1)    (1)
 1  0  8413
 0  1587
So,
1 0
2
Z
  (1)
(1)
1 0 1
(2)  ( 1)  0  9773  0  1587
 0  7186
The Normal Distribution
SUMMARY
 To find (  z ) , the area to the left of a negative
number, we use
(  z )  1  ( z )
N.B. The procedure for this and all other areas involving
negative values can be seen from the diagram.
NEVER try to do these questions without at least
one diagram.