Download The Normal Distribution

The Normal Distribution
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
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Introduction: Normal distribution
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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
μ – 2σ
x
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
μ – 3σ
x
μ + 3σ
99.7% of the distribution lies within
3 standard deviations of the mean.
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Introduction: Normal distribution
68% of young men have a height between 173cm and 189 cm
95% of young men have a height between 165cm and 197 cm
99.7% of young men have a height between 157cm and 205 cm
Only 0.15% of young men have a height taller than 205 cm, or shorter
than 157 cm.
Page 822-823 Exercise 26B Q3, 5, 9, 11, 13
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Introduction: Normal distribution
No matter what the values of  and  are, if it shows a
normal distribution the total area under the curve is 1. We
can consider partial areas under the curve as representing
probabilities.
So in a normal distribution we could find the probability of
P(X < 5) by finding the shaded area in the diagram
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Probabilities using a calculator
Illustrate your answers
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Probabilities using a calculator
Illustrate your answers
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Probabilities using a calculator
Illustrate your answers
b P(X > 194) = 1  P(X< 194)
= 1  0.480061126…..
= 0.520 (to 3 s.f.)
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Probabilities using a calculator
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Page 824-825 Exercise 26C Q3, 5, 6, 7, 8
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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]
It is easy to transform any normal distribution to a
standard normal distribution
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z-distribution
If X ~ N[  ,  ] then Z 
2
N[  ,  ]
2
X 

~ N[0,1].
N[0, 1]
y
y
Standardize
x
-3 -2 -1
1
2
x
3
z-score tells exactly where the score is located relative to
all the other scores in the distribution.
It can also allow us to compare different distributions.
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z-distribution
Yousif scores 89% in an Economics test where the class mean was
72% and the standard deviation was 9.5. He scored 56% in a
Chemistry test where the class average is 45% and the standard
deviation was 6.1. Assuming that the scores for both tests were
normally distributed, in which subject did Yousif score better in
comparison to his class mates.
Economics
z
89  72
 1.79
9.5
Yousif scored 1.79 standard deviations above the mean.
Chemistry
56  45
z
 1.80
6.1
Yousif scored 1.80 standard deviations above the mean.
Yousif scored better in Chemistry than Economics compared
to his class mates.
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Page 829 Exercise 26D
Q2, 5, 6, 7
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The inverse normal function
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The inverse normal function
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The inverse normal function
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The inverse normal distribution
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.
i.e. P(X < x) = 0.14
 = 62 and  = 16
x = 44.7
So, the pass mark is 44.7%.
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Page 831-833 Exercise 26E.1
Q1, 3, 5, 7, 8, 9
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Finding the mean and/or standard deviation
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.
0.025
0.975
μ
0.025
0.975
210
0
1.96
z
Firstly find the z value for 210 grams on your
GDC P(Z > z) = 0.025 using your GDC.
P(Z < z) = 0.975
z = 1.96
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Finding the mean and/or standard deviation
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.
z
1.96 
x

210  

210    1.96
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0.025
0.975
0
1.96
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Finding the mean and/or standard deviation
Secondly, we are told that P(X < 198) = 0.04.
0.04
0.04
198
μ
Find P(Z < z) = 0.04 using your GDC.
–1.75
0
z = -1.75
Solving the two equations simultaneously:
μ = 204 g and
σ = 3.23 g.
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Page 834 Exercise 26E.2
Q1, 2, 6, 7, 8
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