If you observed the heights of people at a busy airport you will
notice that some are short, some are tall, but most would be
around a central value (which would probably depend on their
Gender).
If you displayed a distribution of heights for men and women,
you would get something like:
Male
Heights
Female
Heights
172cm
178cm
Height is an example of a CONTINUOUS RANDOM VARIABLE.
Height is a particular type of random variable whose distribution
Is symmetrical in shape – a sort of bell shape.
This sort of distribution is called the NORMAL DISTRIBUTION
The characteristics of the normal distribution are:
μ
 Symmetrical about the mean, μ
 Mode, median and mean are equal due
to the symmetry of the distribution
 The range of X is from -∞ to +∞
 The horizontal axis is asymptotic
 The total area under the curve is 1
 2/3 of the values lie within 1σ of μ
 95% lie within 2σ of μ
 99.7% lie within 3σ of μ
How does changing μ and σ affect the normal distribution?
μ2 < μ1 < μ3
μ2
μ1
μ3
Changing the mean translates the curve along the x-axis.
It DOES NOT change its shape.
How does changing μ and σ affect the normal distribution?
σ2 < σ1 < σ3
μ
Increasing σ
makes the
curve fatter
and shorter
Decreasing σ
makes the
curve thinner
and taller.
It DOES NOT
MOVE.
If a particular random variable, X, has a normal distribution,
We define the distribution using the notation:
mean
variance
X ~ N(μ, σ2)
If we are finding P(a < X < b), you are
finding an area underneath the curve
a b
STANDARD NORMAL DISTRIBUTION
Situations that demonstrate a Normal distribution will demonstrate
different means and variances and so we need a standardised
model that we can transform each situation to, to allow us to
perform calculations with ease.
For this, we use the STANDARD NORMAL DISTRIBUTION.
It has the properties of any other normal distribution but
Has a mean of 0 and a standard deviation (and variance) of 1.
We give this particular normal distribution the letter Z.
So:
2
Z ~ N(0, 1 )
Probabilities for this distribution are provided in
statistical tables.
Any normal distribution can be transformed onto the standard
normal distribution using the following transformation:
X 
Z

So, if we have the model X ~ N(10, 42) and want to find P(12<X<16)
We can transform the values from X to values on Z by:
16  10 
 12  10
Z
P (12  X  16)  P 
 = P(0.5 < Z < 1.5)
4 
 4
In other words, P(12 < X < 16) on X~ N(10, 42) would be
P(0.5 < Z < 1.5) on Z~N(0, 12)
X ~ N(10, 42)
P(12 < X < 16)
Z ~ N(0, 12)
P(0.5 < Z < 1.5)
0.5 1.5
12 16
P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5)  (1.5)  (0.5)
= 0.9332 - 0.6915
(0.5 )
(1.5 )
1.5
0.5
= 0.2417
If X ~ N(10, 42), what is P(X < 8)?
8  10 

P ( X  8)  P  Z 
 = P(Z < - 0.5)
4 

(0.5 )
( 0.5)
0.5
- 0.5
0.6915
1  (0.5)  ( 0.5)
0.5
P(X < 8) = 1 – 0.6915
= 0.3085
Ex 9A Qu 6
The random variable, A is defined by A ~ N(28, 32). Find:
(a) P(A < 32)
(b) P(A > 36)
(c) P(28 < A < 35)
(d) P(22 < A < 26)
(e) P(25 < A < 33)
32  28 

(a) P(A < 32) = P  Z 
 = P(Z < 1.33)  (1.33) = 0.9082
3 

36  28 

(b) P(A > 36) = P  Z 
 = P(Z > 2.67)  1  (2.67)
3 

= 1 – 0.9962 = 0.0038
(2.67)
2.67
28  28
35  28 

Z
 = P(0 < Z < 2.33)
(c) P(28 < A < 35) = P
3 
 3
 (2.33)  (0) = 0.9901 – 0.5 = 0.4901
22  28
26  28 

Z
 = P(-2 < Z < -0.67)
(d) P(22 < A < 26) = P
3 
 3
 [1  (0.67)]  [1  (2)] = (1-0.7486) – (1-0.9772) = 0.2286
 25  28  Z  33  28 
P
 = P(-1 < Z < 1.67)
(e) P(25 < A < 33) = 
3 
 3
 (1.67)  [1  (1)] = 0.9525 – (1-0.8413) = 0.7938
Ex 9A Qu 13
A machine dispenses liquid into cartons in such a way that the
amount of liquid dispensed on each occasion is normally distributed
with standard deviation 20ml and mean 266ml. Cartons that weigh
less than 260ml have to be recycled. What proportion of cartons are
recycled?
Cartons weighing more than 300ml produce no profit.
What percentage of cartons will this be?
X ~ N(266, 202)
X ~ N(266, 202)
Cartons that weigh less than 260ml have to be recycled.
What proportion have to be recycled?
260  266 

P(X < 260) = P Z 
 = P(Z < -0.3)  1  (0.3)
20


= 1 – 0.6179 = 0.3821 = 38.21%
Cartons weighing more than 300ml produce no profit.
What percentage of cartons will this be?
300  266 

P(X > 300) = P Z 
 = P(Z > 1.7)  1  (1.7)
20


= 1 – 0.9554 = 0.0446 = 4.46%
Ex 9A Qu 15
Over a long period it has been found that the breaking strains of
cables produced by a factory are normally distributed with mean
6000N and standard deviation 150N. Find, to 3 decimal places, the
probability that a cable chosen at random from the production will
have a breaking strain of more than 6200N.
X ~ N(6000, 1502)
6200  6000 

P(X > 6200) = P Z 
 = P(Z > 1.33)  1  (1.33)
150


= 1 – 0.9082 = 0.0918
= 0.092 to 3dp
Ex 9A Qu 18
The thickness of some sheets of wood follows a normal distribution
with mean μ and standard deviation σ. 96% of the sheets will go
through an 8mm gauge while only 1.7% will go through a 7mm
gauge. Find μ and σ.
8  
P Z 
  0.96
0.96
 

8 

μ8
 1.75
8    1.75
-(1)
0.017
7 μ
8    1.75
-(1)
7    2.12 -(2)
(1)-(2)
1 = 3.87σ
σ = 0.258mm
7  

P Z 
  0.017
 

7  
7 

P Z  

0
.
983
 2.12

 


7    2.12 -(2)
Sub σ = 0.258 into (1)
8 – μ = 1.75 x 0.258
μ = 7.5485mm
Packets of breakfast cereal are said to contain 550g. The
manufacturer knows that the weights are normally distributed with
mean 551.2g and standard deviation 15g. What proportion of packets
will contain more than the stated weight?
X ~ N(551.2, 152)
550  551.2 

P(X > 550) = P  Z 

15


= P(Z > -0.08)
 1  ( 0.08)  1  [1  (0.08)]
= 0.5319 = 53.19%
A biologist has studied a particular tropical insect and she has
discovered that its life span is normally distributed. The mean
lifespan of this insect is 72 days and the standard deviation of its
lifespan is eight days. Find the probability that the next insect studied
lives:
(a) Fewer than 70 days
(b) More than 76 days
(c) Between 68 and 78 days.
(a)
X ~ N(72, 82)
70  72 

P(X < 70) = P  Z 
 = P(Z < -0.25)  1  (0.25)
8 

= 1 - 0.5987
= 0.4013
(b)
(c)
X ~ N(72, 82)
76  72 

P(X > 76) = P  Z 
 = P(Z > 0.5)
8 

= 1 - 0.6915
78  72 
 68  72
Z
P(68 < X < 78) = P 

8 
 8
= P(-0.5 < Z < 0.75)
 (0.75)  [1  (0.5)]
= 0.7734 – (1 – 0.6915)
= 0.4649
 1  (0.5)
= 0.3085
The lengths of metal bolts are normally distributed with mean μ and
standard deviation 7cm. If 2.5% of the bolts measure more than
68cm, find the value of μ.
X ~ N(μ, 72)
68   

P Z 
  0.025
P(X > 68) = 0.025
7 

P(Z > 1.96) = 0.025
68  
 1.96
68    13.72
  54.28cm
7
Jam is packed in tins of nominal net weight 1kg. The actual weight of
jam delivered to a tin by the filling machine is normally distributed
about the mean weight set on the machine with a standard deviation
of 12g. The average filling of jam is 1kg.
(a) Calculate the probability that a tin chosen at random contains
less than 985g.
X ~ N(1000, 122)
985  1000 

(a)
P(X < 985) = P  Z 
 = P(Z < -1.25)
12


 1  (1.25)
= 1 - 0.8944 = 0.1056
It is a legal requirement that no more than 1% of tins contain less
than the nominal weight.
(b) Calculate the minimum setting of the filling machine which will
meet this requirement.
X ~ N(μ, 122)
(b)
P(X < 1000) = 0.01
P(Z > 2.3263) = 0.01
1000  
 2.3263
12
1000    27.9156
1000   

P Z 
  0.01
12 

P(Z < -2.3263) = 0.01
  1028g