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AS-Level Maths:
Statistics 1
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S1.6 The normal
distribution
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Contents
Introduction: Normal distribution
Introduction: Normal distribution
The standard normal distribution
More general normal distributions
Solving problems by working backwards
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Introduction: Normal distribution
A sample of heights of 10,000 adult males gave rise to the
following histogram:
Histogram showing the heights of 10000 males
1400
Frequency
1200
1000
800
600
400
200
0
140
148
156
164
172
180
188
More
Height (cm )
Notice that this histogram is symmetrical
and bell-shaped. This is the characteristic
shape of a normal distribution.
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Introduction: Normal distribution
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 normal distribution is an appropriate model for many
common continuous distributions, for example:
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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Introduction: Normal distribution
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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Introduction: Normal distribution
If X has a normal distribution with mean μ, and variance σ2,
we write
X ~ N[μ, σ2]
y
x
μ–σ
μ+σ
68% of the distribution lies within
1 standard deviation of the mean.
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Introduction: Normal distribution
If X has a normal distribution with mean μ, and variance σ2,
we write
X ~ N[μ, σ2]
y
x
μ – 2σ
μ + 2σ
95% of the distribution lies within 2
standard deviations of the mean.
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Introduction: Normal distribution
If X has a normal distribution with mean μ, and variance σ2,
we write
X ~ N[μ, σ2]
y
x
μ – 3σ
μ + 3σ
99.7% of the distribution lies within
3 standard deviations of the mean.
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Introduction: Normal distribution
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:
P(X ≥ 20);
P(–5 < X < 9);
P(X = 19 to the nearest whole number),
i.e. P(18.5 ≤ X < 19.5).
y
Probabilities correspond to areas
underneath the normal curve.
x
There is no simple formula that can be
used to find the probabilities. Instead, the
probabilities are found from tables.
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Contents
The standard normal distribution
Introduction: Normal distribution
The standard normal distribution
More general normal distributions
Solving problems by working backwards
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© Boardworks Ltd 2005
The standard normal distribution
y
x
-3
-2
-1
1
2
3
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.
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The standard normal distribution
Here is an extract from a standard normal distribution table:
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
z
This column
gives
part ofplace
the zofvalue.
row gives
the the
nextfirst
decimal
the z value.
The tables are cumulative, i.e. they give P(Z ≤ z).
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The standard normal distribution
Extract from table:
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
z
P(Z ≤ 0.54) = 0.7054.
So,
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© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
P(Z > 0.6) = 1 – P(Z ≤ 0.6)
= 1 – 0.7257
= 0.2743
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© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
P(0.25 ≤ Z < 0.78) = P(Z ≤ 0.78) – P(Z ≤ 0.25)
= 0.7823 – 0.5987
= 0.1836
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© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
P(Z > –0.3) = P(Z < 0.3)
= 0.6179
16 of 33
Remember that the
standard normal distribution
is symmetrical around 0.
© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
P(Z ≤ –0.28) = 1 – P(Z ≤ 0.28)
= 1 – 0.6103
= 0.3897
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© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
P(–0.08 < Z ≤ 0.85) = P(Z ≤ 0.85) – P(Z ≤ –0.08)
= 0.8023 – (1 – 0.5319)
= 0.3342
18 of 33
© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
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.
19 of 33
© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
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.
20 of 33
© Boardworks Ltd 2005
The standard normal distribution
Extract from table:
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
z
Find c such that P(Z < c) = 0.352.
c must be negative because P(Z < c) is less than 0.5000.
By symmetry, P(Z > |c|) = 0.352 and
P(Z ≤ |c|) = 0.648. Therefore c = –0.38.
21 of 33
© Boardworks Ltd 2005
Contents
More general normal distributions
Introduction: Normal distribution
The standard normal distribution
More general normal distributions
Solving problems by working backwards
22 of 33
© Boardworks Ltd 2005
More general normal distributions
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 ~ N[  ,  ] then Z 
2
N[  ,  2 ]
X 

~ N[0,1].
y
N[0, 1]
y
Standardize
x
x
23 of 33
-3
-2
-1
1
2
© Boardworks Ltd 2005
3
More general normal distributions
Example: If X ~ N[20, 16] , find
a) P(X < 23);
b) P(X > 14);
c) P(16 < X < 24.8).
a) If σ2 = 16, then σ = 4.
Standardize
y
y
23  20
 0.75
4
x
20
23
x
0
0.75
P( X  23)  P( Z  0.75) = 0.7734
24 of 33
© Boardworks Ltd 2005
More general normal distributions
Example: If X ~ N[20, 16] , find
a) P(X < 23);
b) P(X > 14);
c) P(16 < X < 24.8).
b)
Standardize
y
14
20
y
14  20
 1.5
4
x
x
–1.5
0
P( X  14)  P( Z  1.5)  P( Z  1.5) = 0.9332
25 of 33
© Boardworks Ltd 2005
More general normal distributions
Example: If X ~ N[20, 16] , find
a) P(X < 23);
b) P(X > 14);
c) P(16 < X < 24.8).
c) P 16  X  24.8   P  16  20  Z  24.8  20   P(1  Z  1.2)
4
 4

y
y
Standardize
x
16
20
24.8
x
-1
0
1.2
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
26 of 33
© Boardworks Ltd 2005
More general normal distributions
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?
Let X be the random variable for the IQ of an individual.
X ~ N[100, 225].
Standardize
y
y
124  100
 1 .6
15
x
100
124
x
0
1.6
So, we want P(X > 124) = P(Z > 1.6)
= 1 – P(Z ≤ 1.6) = 1 – 0.9452
= 0.0548
27 of 33
© Boardworks Ltd 2005
Contents
Working backwards
Introduction: Normal distribution
The standard normal distribution
More general normal distributions
Solving problems by working backwards
28 of 33
© Boardworks Ltd 2005
Working backwards
Example: If X ~ N[4, 0.25], find the value of
x if P(X < x) = 0.67.
To find x, we start by finding the standardized 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 standardizes to
make 0.44 by rearranging the formula.
Standardize
N[4, 0.25]
y
y
x  4  0.44  0.5
x  4.22
N[0, 1]
x
4
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x
x
0
0.44
© Boardworks Ltd 2005
Working backwards
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.
Let X represent the marks in the examination. X ~ N[62, 256].
We need to find x such that P(X ≥ x) = 0.86.
x  62
We need to solve:
 1.08. Therefore x = 44.72.
16
y
Standardize
y
x
x
x
–1.08
So, the pass mark is 44.
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© Boardworks Ltd 2005
Working backwards
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.
We are given 2 pieces of information which we can
use to form equations:
210  
Firstly:
 1.96  210    1.96

0.975
μ
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0.975
0.025
210
0
1.96
© Boardworks Ltd 2005
Working backwards
Secondly, we are told that P(X < 198) = 0.04.
This gives the equation:
198  

0.04
 1.75  198    1.75
0.96
0.96
198
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μ
–1.75
0
© Boardworks Ltd 2005
Working backwards
The two equations are:
210    1.96
198    1.75
Subtracting to eliminate μ:
12  3.71    3.2345 g
This gives μ = 210 – 1.96 × 3.2345 = 203.66 g
So the solutions to 3 s.f. are μ = 204 g and σ = 3.23 g.
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© Boardworks Ltd 2005