Download the normal distribution - Mathematics with Mr Walters

S1 STATISTICS
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THE NORMAL DISTRIBUTION
This is a model for continuous random variables.
P(a < X < b)
a
b
x
The graph is symmetrical about the mean μ
Therefore, mode = median = mean
Total area under the curve = 1 (unity)
The shaded area gives the probability that X is between a
and b
We say that the random variable X is distributed normally
with mean value μ and variance σ2 (these are the parameters of
the distribution)
This statement is represented by the notation:
X ~ N (μ, σ2)
E[X] = μ
Var[X] = σ2
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These two distributions have different locations (means) but
the same the same dispersion (variance):-
X
X
1
c
2
d
These two distributions have the same location (mean) but
different dispersions (variances):X
1
X2
c
Note:- Total area under all curves = 1
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Returning to the area under the curve needed to calculate
probabilities:-
Standard techniques (integration) cannot be used to evaluate
this area.
Statistical tables have to be referred to.
It would require an infinite number of such tables to cope with
all possible values for the mean and variance, so all such
distributions are compared with one special one:-
THE STANDARD NORMAL DISTRIBUTION
This has mean value 0 and variance 1
It is represented by the letter Z
Z ~ N (0, 1)
0
Φ (z) = P (Z < z)
z
Z
this is the Cumulative Distribution
Function (Φ = Phi pronounced “fie”)
The table on p 201 gives the values of the Normal Distribution
Function
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HOW TO READ THE TABLES
Always use a sketch
Examples
1)
P (Z < 0.72)
= 0.7642
Z
0.72
2)
P (Z > 0.72)
= 1 – P(Z < 0.72)
= 1 – 0.7642
= 0.2358
Z
0.72
3)
P (Z > − 0.6)
= P(Z < 0.6)
by symmetry, this
is the same as
-0.6
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Z
= 0.7257
0.6
Z
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4)
SLBS
P (Z < − 1.2)
= P(Z > 1.2)
= 1 – P(Z < 1.2)
symmetry =>
= 1 – 0.8849
= 0.1151
5)
1.2
Z
-1.2
Z
P (− 0.4 < Z < 1.75)
= P(Z < 1.75) – P(Z < – 0.4)
= P(Z < 1.75) – [1 – P(Z < 0.4)]
= 0.9599 – 1 + 0.6554
= 0.6153
-0.4
1.75
Z
Ex 9A p 179
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USING TABLES TO FIND z GIVEN THE PROBABILITY
The tables can be read in reverse!
If you know that P (Z < k) = 0.8888
Look for 0.8888 in the main table in one of the P(Z<z) columns
The corresponding value of z gives the value of k
z
P(Z<z)
1.22 0.8888
Hence, k = 1.22
Example
P(Z<z) = 0.3 find z
Tables only give values of P(Z<z) from 0.5 up to 1
So z will be a negative quantity (see Example 4)
P(Z<−z) = 1 − P(Z<z) = 0.7
Tables give −z = 0.52 hence z = − 0.52
PERCENTAGE POINTS TABLE
This is another table that can be used and is very popular with
the examiners! (on p202)
This reads the main table backwards!
It gives the values of z in the main table for which Z exceeds
with probability p
i.e. P(Z>z) = 1 − P(Z<z) = p
Whenever possible this table should be used to find a value of
z given a value for p = P(Z>z)
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Example
If 30% of the values of Z are greater than z
Then z = 0.5244
(This is found by looking up 0.3 under the column headed p and
reading off the corresponding value of z)
If told that 70% of the values of Z are greater than z
Then z = − 0.5244
(This is found by considering symmetry)
Note: this table enables you to deal only with certain
percentages:50%
40% (+ 60%)
30% (+70%)
20% (+80%)
15% (+85%)
10% (+90%)
5% (+95%)
2.5% (+97.5%)
1% (+99%)
0.5% (+99.5%)
0.1% (+99.9%)
0.05% (+99.95%)
Ex 9B p 181
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STANDARDISING A NORMAL DISTRIBUTION
Any Normal Distribution can be mapped onto the Standard
Normal Distribution by a simple method of coding.
Subtracting the mean value shifts the axis of symmetry from
z = μ to the value where z = 0
Dividing by the standard deviation (σ) reduces the variance to 1
If X ~ N(μ, σ2)
Then Z =
X – 

where Z ~ N(0, 1)
Given the mean and the variance, you can find the probability
that X lies within any given range.
Example
2
If X ~ N(14,2 ) find P(13 < X < 16)
 13 – 14
16 – 14 
= P 
< Z <

2
2


= P( – 0·5 < Z < 1)
= P(Z < 1) – P(Z < – 0.5)
= P(Z < 1) – [1 – P(Z < 0.5)]
= 0.8413 – [1 – 0.6915]
= 0.5328
-0.5
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Z
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SOME STANDARD RESULTS
For any normal distribution of X, the probability that X lies
within 1 standard deviation of the mean can be found:
X can be mapped onto Z
0
Z
P( – 1 < Z < 1) = P(Z < 1) – P(Z < – 1)
= P(Z < 1) – P(Z > 1)
= P(Z < 1) – [1 – P(Z < 1)]
= 2P(Z < 1) – 1
= 2  0·8413 – 1
= 0·6826
 68%
Similarly
0
Z
0
Z
P(within 2 s.d. of mean) = 2P(Z < 2) – 1
= 0·9544  95%
P(within 3 s.d. of mean) = 2P(Z < 3) – 1
= 0·9974  99·7%
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If you are given the mean and variance and the probability that
X is less than or greater than an unknown value, you can find
that unknown value.
Example
X has a normal distribution with mean 5 and variance 0.81.
Find x, such that P(X < x) = 0.01
X ~ N(5, 0.92)

x – 5 
P(X < x) = P  Z <

0·9 

= 0·01
=
0.01
x – 5
0
0.01
X
0
0.9
X
5 – x
0.9
Tables only give values of p(Z<z) from 0.5 up to 1

x – 5 

5 – x 
Symmetry  P  Z > –
=
P
Z
>


 = 0·01
0·9 
0·9 



5 – x 
 P  Z <
 = 0·99
0·9 

and tables 
5 – x
0·9
= 2·32
 5 – x = 2·32  0·9  x = 5 – 2·32  0·9
= 2·912
Ex 9C p 184
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FINDING THE MEAN AND/OR THE VARIANCE
The most difficult problems are when you are given the
probability that X is less than or greater than a specific value
together with the mean (or variance) and you have to calculate
the variance (or mean) of the distribution.
Example
X is normally distributed with standard deviation 10
P(X > 40) = 0.25. Find the mean of the distribution
X ~ N(μ, 102)

40 –  
P(X > 40) = P  Z >
 = 0·25



0.25
40 – 
X
10

40 –  
P  Z <
 = 0·75
10


40 – 
10
= 0·675
40 –  = 0·675  10 = 6·75
 = 40 – 6·75
 = 33·25
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If you have to find both the mean and the variance, you will
need two pieces of information and use simultaneous equations
Example
X is normally distributed
P(X > 76) = 0.0354 and P(Z < 47) = 0.1
Find the mean and the variance.

P(X > 76) = P  Z >


0.1
0.0354
P  Z <

76 – 

 = 0.0354

76 –  

 = 1 – 0.0354 = 0.9646


Tables 
47 –  0
76 –  
76 – 

= 1.81  76 –  = 1.81 (1)
X


47 –  
P(X < 47) = P  Z <
 = 0·1




 47 –   

 47 –   
=
0·1
P
Z
<
–
P  Z > – 

 


  = 0·9




 


 

 – 47 
P  Z <
 = 0·9



Tables 
 – 47

= 1·28   – 47 = 1·28 (2)
(1) + (2)  29 = 3·09   = 9·39 to 2dp
Subst in (2)   = 59·01 to 2dp
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Ex 9D p 188
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PROBLEMS IN CONTEXT
Example
A lift states that its maximum safe load is 90kg. The weights
of people using the building are normally distributed with mean
67kg and variance 60 kg2. The lift is meant for books and
equipment only. What is the probability that someone who is
tempted to use the lift exceeds the safety limit?
X ~ N (67, 60)
=>
σ = √60
67
90
X

90 – 67 
P(X > 90) = P  Z >

60


= P(Z > 2·97)
= 1 – P(Z < 2·97)
= 1 – 0·9984
= 0·0016
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Example
On a farm, the middle 80% of the eggs produced have masses
between 45g and 70g. Assuming that the masses of the eggs
are normally distributed, find their mean and standard
deviation.
10%
10%
70
45
Symmetry   =
45 + 70
2
X
= 57·5 g
P(X < 70) = 0·9

70 – 57·5 
P  Z <
 = 0·9



Tables 
70 – 57·5

= 1·28
70 – 57·5 = 1·28
12·5 = 1·28
 = 9·77g to 2dp
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Example
The mass of beans in a tin is normally distributed with mean
250g and standard deviation 10g. It is a legal requirement that
no more than 5% of the tins should have a mass less than that
stated on the label. What mass should the label state?
X ~ N(250, 102)
5%
W
250
X

W – 250 

250 – W 
P(X < W) = P  Z <
=
P
Z
>


 = 0·05
10
10




Percentage Points table 
250 – W
10
= 1·6449
 W = – (1·6449  10) + 250 = 233·551
 The label should read 233g
Mixed Ex 9D
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