THE NORMAL DISTRIBUTION Lesson 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 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) = a

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) =

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) = This is a really useful technique 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)

P(Z<2.16) =

Ex 4 P(-1.24<Z<2.16) P(Z<2.16) = P(Z<-1.24) = 1-P(Z<1.24) = =

Ex 4 P(-1.24<Z<2.16) P(Z<2.16) = P(Z<-1.24) = 1-P(Z<1.24) = = P(-1.24<Z<2.16) = – =

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

Blank Normal Distribution Plot