THE NORMAL DISTRIBUTIONLesson 1

Objectives• To introduce the normal distribution• The standard normal distribution• Finding probabilities under the curve

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…

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

Find P (Z≤a)To do this we begin with a sketch of the normal distribution

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

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

Ex1 Find P (Z<1.55)To do this we begin with a sketch of the normal distribution

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

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

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)

The Normal Distribution Table

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

Ex 2 Find P(Z>1.74)

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

Ex 3 P(Z<-0.83)As our table only has values for the positive side of the distribution we must use symmetry…

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)

This is a really useful result

P(Z<-a) = 1 - P(Z<a)

Ex 4 P(-1.24<Z<2.16)

Ex 4 P(-1.24<Z<2.16)

P(Z<2.16) = 0.9846

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

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

Recommended work…• Read through pages 177 and 178.• Do Exercise 9A p179

Blank Normal Distribution Plot

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