normal-distribution-2.ppt

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AS-Level Maths:
Statistics 1
for Edexcel
S1.6 The normal
distribution

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Contents
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Introduction: Normal distribution
The standard normal distribution
More general normal distributions
Solving problems by working backwards.

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Histogram showing the heights of 10000 males
0
200
400
600
800
1000
1200
1400
140 148 156 164 172 180 188 More
Height (cm)
Frequency
A sample of heights of 10,000 adult males gave rise to the
following histogram:
Notice that this histogram is symmetrical
and bell-shaped. This is the characteristic
shape of a normal distribution.

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The normal distribution is an appropriate model for many
common continuous distributions, for example:
If we were to draw a
smooth curve through the
mid-points of the bars in
the histogram of these
heights, it would have the
following shape:
This is called the
normal curve.
The masses of new-born babies;
The IQs of school students;
The hand span of adult females;
The heights of plants growing in a field;
etc.

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All normal curves are symmetrical and bell-shaped but the
exact shape is governed by 2 parameters – the mean, μ, and
the standard deviation, σ.

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If X has a normal distribution with mean μ, and variance σ2,
we write
X ~ N[μ, σ2]
68% of the distribution lies within
1 standard deviation of the mean.
x
y
μ – σ μ + σ

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95% of the distribution lies within 2
standard deviations of the mean.
x
y
we write
X ~ N[μ, σ2]
μ – 2σ μ + 2σ

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99.7% of the distribution lies within
3 standard deviations of the mean.
x
y
we write
X ~ N[μ, σ2]
μ + 3σ
μ – 3σ

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As normal distributions always represent continuous data, it
only makes sense to find the probability that X takes a value in
a particular interval. For example, we could find:
There is no simple formula that can be
used to find the probabilities. Instead, the
probabilities are found from tables.
Probabilities correspond to areas
underneath the normal curve.
P(X ≥ 20);
P(–5 < X < 9);
P(X = 19 to the nearest whole number),
i.e. P(18.5 ≤ X < 19.5).
x
y

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Contents
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The normal distribution with mean 0 and standard deviation 1
is called the standard normal distribution – it is denoted Z.
So, Z ~ N[0, 1].
Probabilities for this distribution are given in tables.
-3 -2 -1 1 2 3
x
y

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Here is an extract from a standard normal distribution table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
The tables are cumulative, i.e. they give P(Z ≤ z).
This column gives the first part of the z value.
This row gives the next decimal place of the z value.

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So, P(Z ≤ 0.54) = 0.7054.
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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P(Z > 0.6) = 1 – P(Z ≤ 0.6)
= 1 – 0.7257
= 0 .2743
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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P(0.25 ≤ Z < 0.78) = P(Z ≤ 0.78) – P(Z ≤ 0.25)
= 0.7823 – 0.5987
= 0.1836
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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P(Z > –0.3) = P(Z < 0.3)
= 0.6179
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
Remember that the
standard normal distribution
is symmetrical around 0.

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P(Z ≤ –0.28) = 1 – P(Z ≤ 0.28)
= 1 – 0.6103
= 0.3897
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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P(–0.08 < Z ≤ 0.85) = P(Z ≤ 0.85) – P(Z ≤ –0.08)
= 0.8023 – (1 – 0.5319)
= 0.3342
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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Find a such that P(Z < a) = 0.6950.
We search in the table to find the probability 0.6950.
We see that a = 0.51.
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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Find b such that P(Z > b) = 0.242.
i.e. such that P(Z ≤ b) = 1 – 0.242 = 0.758.
We see that b = 0.7.
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

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Find c such that P(Z < c) = 0.352.
c must be negative because P(Z < c) is less than 0.5000.
Extract from table:
z 0 1 2 3 4 5 6 7 8 9
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
By symmetry, P(Z > |c|) = 0.352 and
P(Z ≤ |c|) = 0.648. Therefore c = –0.38.

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Contents
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It would of course be impractical to publish tables of
probabilities for every possible normal distribution.
Fortunately, it is possible and easy to transform any
normal distribution to a standard normal:
If X ~ then ~ [ ].
N 0,1
X
Z




[ , ]
2
N  
Standardise
[ ]
N 0, 1
x
y
[ , ]
2
N  
-3 -2 -1 1 2 3
x
y

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Example: If , find
a) P(X < 23);
b) P(X > 14);
c) P(16 < X < 24.8).
~ [ , ]
N 20 16
X
a) If σ2 = 16, then σ = 4.
( ) ( . )
P 23 P 0 75
X Z
  
x
y
x
y
20 23 0 0.75
Standardise
.
23 20
0 75
4


= 0.7734

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b)
( ) ( . ) ( . )
P 14 P 1 5 P 1 5
X Z Z
     
Example: If , find
a) P(X < 23);
b) P(X > 14);
c) P(16 < X < 24.8).
~ [ , ]
N 20 16
X
x
y
x
y
14 20 –1.5 0
Standardise
.
14 20
1 5
4

 
= 0.9332

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c)  
.
. ( . )
16 20 24 8 20
P 16 24 8 P P 1 1 2
4 4
X Z Z
 
 
        
 
 
P(Z < 1.2) = 0.8849
and P(Z < –1) = 1 – P(Z < 1) = 1 – 0.8413 = 0.1587.
So, P(–1 < Z < 1.2) = 0.8849 – 0.1587 = 0.7262
Example: If , find
a) P(X < 23);
b) P(X > 14);
c) P(16 < X < 24.8).
~ [ , ]
N 20 16
X
x
y
x
y
Standardise
16 20 24.8 -1 0 1.2

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Let X be the random variable for the IQ of an individual.
X ~ N[100, 225].
So, we want P(X > 124) = P(Z > 1.6)
= 1 – P(Z ≤ 1.6) = 1 – 0.9452
Examination style question: IQs are normally distributed
with mean 100 and standard deviation 15. What proportion
of the population have an IQ of at least 124?
x
y
x
y
100 124 0 1.6
Standardise
.
124 100
1 6
15


= 0.0548

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Contents
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Working backwards

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x
y
To find x, we start by finding the standardised value z such
that P(Z < z) = 0.67.
From tables we see that z = 0.44.
We therefore need to find the value that standardises to
make 0.44 by rearranging the formula.
Example: If X ~ N[4, 0.25], find the value of
x if P(X < x) = 0.67.
Working backwards
x
y
4 x 0 0.44
Standardise
[ . ]
N 4, 0 25 [ ]
N 0, 1
.
.
.
4 0 44 0
4 22
5
x
x
  


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x
y
x
y
Let X represent the marks in the examination. X ~ N[62, 256].
We need to find x such that P(X ≥ x) = 0.86.
We need to solve: . Therefore x = 44.72.
Example: Marks in an examination can be assumed to
follow a normal distribution with mean 62 and standard
deviation 16. The pass mark is to be chosen so that 86%
of candidates pass. Find the pass mark.
-1.08
.
62
1 08
16
x 
 
x
Working backwards
Standardise
So, the pass mark is 44.

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Examination style question: A machine is designed to fill jars
of coffee so that the contents, X, follow a normal distribution
with mean μ grams and standard deviation σ grams.
If P(X > 210) = 0.025 and P(X < 198) = 0.04, find μ and σ
correct to 3 significant figures.
μ 210
0.975
0 1.96
. .
210
1 96 210 1 96

 


   
0.025
Working backwards
We are given 2 pieces of information which we can
use to form equations:
Firstly:
0.975

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Secondly, we are told that P(X < 198) = 0.04.
This gives the equation:
198 μ
0.04
-1.75 0
. .
198
1 75 198 1 75

 


    
0.96
Working backwards
0.96

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The two equations are:
.
198 1 75
 
 
.
210 1 96
 
 
Subtracting to eliminate μ:
. .
12 3 71 3 2345g
 
  
This gives μ = 210 – 1.96 × 3.2345 = 203.66g
So the solutions to 3 s.f. are μ = 204g and σ = 3.23g.
Working backwards

normal-distribution-2.ppt

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