Report

THE NORMAL DISTRIBUTION Lesson 2 Starter: Find P (Z<1.63) To do this we begin with a sketch of the normal distribution. We then mark a line to represent Z=1.63 P(Z<1.63) is the area under the curve to the left of a. We now use the table to look up this probability 1.63 P(Z<1.63) = 0.9484 Remember this from last lesson! Note this result.. P(Z>a) = 1-P(Z<a) And we found this really useful result P(Z<-a) = 1 - P(Z<a) Objectives • Finding the value of Z from a given probability. Points to note • P(Z<a) is greater than 0.5 •a>0 • P(Z<a) is less than 0.5 • a <0 Points to note • P(Z>a) is less than 0.5 •a>0 • P(Z>a) is greater than 0.5 •a<0 Ex 1 Find the value of a such that P(Z<a)=0.7852 Ex 1 Find the value of a such that P(Z<a)=0.7852 Use the table P(Z<a) = 0.7852 a = 0.79 a Ex 2 Find the value of a such that P(Z>a)=0.01 P(Z>a) = 0.01 P(Z<a) = 0.990 Look up this result is in the main table. It’s not there! Instead look at the table of Percentage Points to see if p=0.01 is listed. a P(Z>a) = 0.01 a = 2.3263 Ex 3 Find the value of a such that P(Z>a)=0.0314 P(Z>a) = 0.0314 P(Z<a) = 1 – P(Z>a) = 1-0.0314 = 0.9686 So from the main table a = 1.86 a a Ex 4 Find the value of a such that P(Z<a) = 0.0197 The table only lists values for z>0 so we need to reflect the problem in the vertical axis… a p=0.0197 P(Z<z) = 1 – 0.0197 = 0.9803 From the table z = 2.06 Therefore a = -2.06 z p=0.9830 Further Learning: • Read through Example 2 on P180 • Do Ex 9B on P181