the normal distribution lesson 1. objectives to introduce the normal distribution the standard...
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THE NORMAL DISTRIBUTIONLesson 1
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Objectives• To introduce the normal distribution• The standard normal distribution• Finding probabilities under the curve
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Normal Distributions
Normal distributions are used to model continuous variables in many different situations.
For example, a normal distribution could be used to model the height of students.
We can transform our normal distribution into a standard normal distribution…
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The standard normal variable
The standard normal variable is usually denoted by Z
It has a mean =0 and a standard deviation of =1
)
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The curve is designed so that the total area underneath the curve is 1
The curve fits within ±4 standard deviations from the mean
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Find P (Z≤a)To do this we begin with a sketch of the normal distribution
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P (Z≤a)To do this we begin with a sketch of the normal distribution.
We then mark a line to represent Z=a
a
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P (Z≤a)To do this we begin with a sketch of the normal distribution.
We then mark a line to represent Z=a
P(Z≤a) is the area under the curve to the left of a.
For continuous distributions there is no difference between P(Z≤a) and P(Z<a)
a
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Ex1 Find P (Z<1.55)To do this we begin with a sketch of the normal distribution
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Ex 1 Find P (Z<1.55)To do this we begin with a sketch of the normal distribution.
We then mark a line to represent Z=1.55
a
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Ex 1 Find P (Z<1.55)To do this we begin with a sketch of the normal distribution.
We then mark a line to represent Z=a
P(Z<1.55) is the area under the curve to the left of a.
We now use the table to look up this probability
a
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The Normal Distribution Table• The table describes the
positive half of the bell shaped curve…
• (Z) is sometimes used as shorthand for P(Z<z)
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The Normal Distribution Table
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Ex 1 Find P (Z<1.55)To do this we begin with a sketch of the normal distribution.
We then mark a line to represent Z=a
P(Z<1.55) is the area under the curve to the left of a.
We now use the table to look up this probability
P(Z<1.55) = 0.9394
a
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Ex 2 Find P(Z>1.74)
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Ex 2 Find P(Z>1.74)
Note this result..
P(Z>a) = 1-P(Z<a)
So
P(Z>1.74) = 1 – P(Z<1.74) = 1 – (0.9591) = 0.0409
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Ex 3 P(Z<-0.83)As our table only has values for the positive side of the distribution we must use symmetry…
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Ex 3 P(Z<-0.83)We have reflected the curve in the vertical axis.
P(Z<-0.83) = 1- P(0.83) = 1 – (0.7967) = 0.2033
This is a really useful technique
P(Z<-a) = P(Z>a)= 1- P(Z<a)
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This is a really useful result
P(Z<-a) = 1 - P(Z<a)
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Ex 4 P(-1.24<Z<2.16)
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Ex 4 P(-1.24<Z<2.16)
P(Z<2.16) = 0.9846
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Ex 4 P(-1.24<Z<2.16)
P(Z<2.16) = 0.9846
P(Z<-1.24) = 1-P(Z<1.24) = 1-0.8925 = 0.1075
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Ex 4 P(-1.24<Z<2.16)
P(Z<2.16) = 0.9846
P(Z<-1.24) = 1-P(Z<1.24) = 1-0.8925 = 0.1075
P(-1.24<Z<2.16) = 0.9846 – 0.1075= 0.8771
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Recommended work…• Read through pages 177 and 178.• Do Exercise 9A p179
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Blank Normal Distribution Plot
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