Lesson #15

The NormalDistribution

For a truly continuous random variable,

P(X = c) = 0for any value, c.

Thus, we define probabilities only on intervals.

P(X < a)

P(X > b)

P(a < X < b)

f(x) is the probability density function, pdf.

This gives the height of the “frequency curve”.

Probabilities are areas under the frequency curve!

f(x) is the probability density function, pdf.

This gives the height of the “frequency curve”.

Probabilities are areas under the frequency curve!

Remember this!!!

a b

P(X < a)

P(a < X < b)

P(X > b)

P(X < a) = P(X < a) = F(a)

x

f(x)

P(X > b) = 1 - P(X < b) = 1 - F(b)

P(a < X < b) = P(X < b) - P(X < a) = F(b) - F(a)

If X follows a Normal distribution, withparameters and 2, we use the notation

X ~ N( , 2)

E(X) = Var(X) = 2

2

x - 21

f(x) = e - < x < 2

2

-

A standard Normal distribution is one where = 0 and 2 = 1. This is denoted by Z

Z ~ N()

-3 -2 -1 0 1 2 3

Table A.3 in the textbook gives upper-tailprobabilities for a standard Normal distribution,and only for positive values of Z.

-3 -2 -1 0 1 2 3

P(Z > 1)

Table C in the notebook gives cumulativeprobabilities, F(x), for a standard Normaldistribution, for –3.89 < Z < 3.89.

-3 -2 -1 0 1 2 3

P(Z < -1)

P(Z < 1.27)

-3 -2 -1 0 1 2 3

1.27

= .8980

P(Z < -0.43)

-3 -2 -1 0 1 2 3

-0.43

= .3336

0.43

= P(Z > 0.43)

P(Z > -0.22)

-3 -2 -1 0 1 2 3

-0.22

= 1 – P(Z < -0.22)= 1 – .4129 = .5871

P(-1.32 < Z < 0.16)

-3 -2 -1 0 1 2 3

-1.32

= .5636 – .0934 = .4702= P(Z < 0.16) - P(Z < -1.32)

0.16

Find c, so that P(Z < c) = .0505

-3 -2 -1 0 1 2 3

.0505

c

c = -1.64

Find c, so that P(Z < c) .9

-3 -2 -1 0 1 2 3

.9

c

c 1.28

Find c, so that P(Z > c) = .166

-3 -2 -1 0 1 2 3

.166

c

c = 0.97 P(Z < c) = 1 - .166 = .834

.834

ZP is the point along the N(0,1) distributionthat has cumulative probability p.

-3 -2 -1 0 1 2 3

p

ZP

Z.0505 = -1.64

Z.9 1.28

Z.975 = 1.96