the normal probability distribution points of inflection 2 3 2 3 2 3 2 3

The Normal Probability Distribution

Points of Inflection

2 3 2 3

2 3 2 3

Main characteristics of the Normal Distribution

• Bell Shaped, symmetric

• Points of inflection on the bell shaped curve are at – and + That is one standard deviation from the mean

• Area under the bell shaped curve between – and + is approximately 2/3.

• Area under the bell shaped curve between – 2 and + 2is approximately 95%.

• Close to 100% of the area under the bell shaped curve between – 3 and + 3

There are many Normal distributions

depending on by and

0

0.01

0.02

0.03

0 50 100 150 200

x

f(x)

0

0.01

0.02

0.03

0 50 100 150 200

x

f(x)

0

0.01

0.02

0.03

0 50 100 150 200

x

f(x)

Normal = 100, = 40 Normal = 140, =20

Normal = 100, =20

The Standard Normal Distribution = 0, = 1

0

0.1

0.2

0.3

0.4

-3 -2 -1 0 1 2 3

• There are infinitely many normal probability distributions (differing in and )

• Area under the Normal distribution with mean and standard deviation can be converted to area under the standard normal distribution

• If X has a Normal distribution with mean and standard deviation than has a standard normal distribution

has a standard normal distribution.

• z is called the standard score (z-score) of X.

X

z

Converting Area

under the Normal distribution with mean and standard deviation

to

Area under the standard normal distribution

Perform the z-transformation

then

Area under the Normal distribution with mean and standard deviation

X

z

P a X b

a X bP

a bP z


P a X b

Area under the Normal distribution with mean and standard deviation

a b

a bP z


a b

0

1

Using the tables for the Standard Normal distribution

Example

Find the area under the standard normal curve between z = - and z = 1.45

9265.0)45.1( zP

• A portion of Table 3:

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06

1.4 0.9265

...

...

0 145.

9265.0

z0 145.

9265.0

z

P z( 0. ) 98 .01635

Example

Find the area to the left of -0.98; P(z < -0.98)

0000.98

Area asked for

0.980.98

Area asked for

0735.09265.00000.1)45.1( zP

Example

Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)

9265.0

145.

Area asked for

0 z

9265.0

145.

Area asked for

145.

Area asked forArea asked for

0 z0 z

4265.05000.09265.0)45.1( zP

Example

Find the area to the between z = 0 and of z = 1.45; P(0 < z < 1.45)

• Area between two points = differences in two tabled areas

145.0 z145.145.0 z0 z

Notes

Use the fact that the area above zero and the area below zero is 0.5000

the area above zero is 0.5000

When finding normal distribution probabilities, a sketch is always helpful

3962.01038.05000.0)026.1( zP

Example:Find the area between the mean (z = 0) and z = -1.26

0 z 1 2 6.

A r e a a s k e d f o r

0 z0 z 1 2 6.


1 2 6.

A r e a a s k e d f o rA r e a a s k e d f o r

Example: Find the area between z = -2.30 and z = 1.80

9534.00107.09641.0)80.126.1( zP

0 .. - 2 . 3 0

Area Required

1 . 8 00 .. 0 .. 0 .. - 2 . 3 0

Area Required

1 . 8 0- 2 . 3 0

Area Required Area Required

1 . 8 0

Example: Find the area between z = -1.40 and z = -0.50

2277.00808.03085.0)50.040.1( zP

Area asked for

-1.40 -0.500

Area asked for

-1.40 -0.50

Area asked for

-1.40 -0.5000

Computing Areas under the general Normal Distributions

(mean , standard deviation )

1. Convert the random variable, X, to its z-score.

Approach:

3. Convert area under the distribution of X to area under the standard normal distribution.

2. Convert the limits on random variable, X, to their z-scores.

X

z

b

za

PbXaP

Example

Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally

distributed and a bottle is selected at random:

1) Find the probability the bottle contains between 32.00 oz and 32.025 oz

2) Find the probability the bottle contains more than 31.97 oz

When x z 32.0032.00 32.00 32.0

0.00;0.02

Solutions part 1)

When x z

32 025

32.025 32 025 32.0125. ;

.

0.02.

P X PX

P z

( . )0. 0.

.

0.

( . ) .

32.0 32 02532.0 32.0

02

32.0

02

32 025 32.0

02

0 125 0 3944

Graphical Illustration:

3 2 . 0 x0 1 2 5. z

3 2 0 2 5.


3 2 . 0 x3 2 . 0 x0 1 2 5. z0 1 2 5. z

3 2 0 2 5.


3 2 0 2 5.


P x Px

P z( . ).

( .

. . .

319732.0

0.023197 32.0

0.02150)

1 0000 00668 0 9332

Example, Part 2)

32.03197. x

0150. z32.03197. x

0150.150. z

Combining Random Variables

Quite often we have two or more random variables

X, Y, Z etc

We combine these random variables using a mathematical expression.

Important question

What is the distribution of the new random variable?

An Example

Suppose that a student will take three tests in the next three days

1. Mathematics (X is the score he will receive on this test.)

2. English Literature (Y is the score he will receive on this test.)

3. Social Studies (Z is the score he will receive on this test.)

Assume that

1. X (Mathematics) has a Normal distribution with mean = 90 and standard deviation = 3.

2. Y (English Literature) has a Normal distribution with mean = 60 and standard deviation = 10.

3. Z (Social Studies) has a Normal distribution with mean = 70 and standard deviation = 7.

Graphs

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 20 40 60 80 100

X (Mathematics) = 90, = 3.

Y (English Literature) = 60, = 10.

Z (Social Studies) = 70 , = 7.

Suppose that after the tests have been written an overall score, S, will be computed as follows:

S (Overall score) = 0.50 X (Mathematics) + 0.30 Y (English Literature) + 0.20 Z (Social Studies) + 10 (Bonus marks)

What is the distribution of the overall score, S?

Sums, Differences, Linear Combinations of R.V.’s

A linear combination of random variables, X, Y, . . . is a combination of the form:

L = aX + bY + …

where a, b, etc. are numbers – positive or negative.

Most common:Sum = X + Y Difference = X – Y

Others

Averages = 1/3 X + 1/3 Y + 1/3 Z

Weighted averages = 0.40 X + 0.25 Y + 0.35 Z

Means of Linear Combinations

The mean of L is:

Mean(L) = a Mean(X) + b Mean(Y) + …

L = a X + b Y + …

Most common:

Mean( X + Y) = Mean(X) + Mean(Y)

Mean(X – Y) = Mean(X) – Mean(Y)

If L = aX + bY + …

Variances of Linear Combinations

If X, Y, . . . are independent random variables and

L = aX + bY + … then

Variance(L) = a2 Variance(X) + b2 Variance(Y) + …

Most common:

Variance( X + Y) = Variance(X) + Variance(Y)

Variance(X – Y) = Variance(X) + Variance(Y)

2 2 2 2 2L X Ya b

If X, Y, . . . are independent normal random variables, then L = aX + bY + … is normally distributed.

In particular:

X + Y is normal with

X – Y is normal with

Combining Independent Normal Random Variables

22 deviation standard

mean

YX

YX

22 deviation standard

mean

YX

YX

Example: Suppose that one performs two independent tasks (A and B):

X = time to perform task A (normal with mean 25 minutes and standard deviation of 3 minutes.)

Y = time to perform task B (normal with mean 15 minutes and std dev 2 minutes.)

X and Y independent so T = X + Y = total time is normal with

6.323 deviation standard

401525 mean

22

0823.39.16.3

404545

ZPZPTP

What is the probability that the two tasks take more than 45 minutes to perform?

The distribution of averages (the mean)

• Let x1, x2, … , xn denote n independent random variables each coming from the same Normal distribution with mean and standard deviation .

• Let

11 2

1 1 1

n

ii

n

xx x x x

n n n n

What is the distribution of ?x

The distribution of averages (the mean)

Because the mean is a “linear combination”

1 2

1 1 1nx x x xn n n

and

1 1 1 1n

n n n n

1 2

2 2 22 2 2 21 1 1

nx x x xn n n

2 2 2 2 22 2 2

2

1 1 1n

n n n n n

Thus if x1, x2, … , xn denote n independent random variables each coming from the same Normal distribution with mean and standard deviation .

Then

11 2

1 1 1

n

ii

n

xx x x x

n n n n

has Normal distribution with

mean andx 2

2variance x n

standard deviation xn

Example

• Suppose we are measuring the cholesterol level of men age 60-65

• This measurement has a Normal distribution with mean = 220 and standard deviation = 17.

• A sample of n = 10 males age 60-65 are selected and the cholesterol level is measured for those 10 males.

• x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, are those 10 measurements

Find the probability distribution of

Compute the probability that is between 215 and 225

?xx

Solution

Find the probability distribution of x

Normal with 220x 17

and 5.37610

xn

215 225P x

215 220 220 225 220

5.376 5.376 5.376

xP

0.930 0.930 0.648P z

Graphs

0

0.02

0.04

0.06

0.08

150 170 190 210 230 250 270 290 310

The probability distribution of individual observations

The probability distribution of the mean

Normal approximation to the Binomial distribution

Using the Normal distribution to calculate Binomial probabilities

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

-

-0.5

Binomial distribution

Approximating

Normal distribution

Binomial distribution n = 20, p = 0.70

049.2

14

npq

np

Normal Approximation to the Binomial distribution

• X has a Binomial distribution with parameters n and p

21

21 aYaPaXP

• Y has a Normal distribution

npq

np

correction continuity21

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

Binomial distribution

-

0.0500

0.1000

0.1500

0.2000

0.2500

a

-

-0.5

Approximating

Normal distribution

P[X = a]

21a 2

1a

-

0.0500

0.1000

0.1500

0.2000

0.2500

a-

-0.5

21

21 aYaP

-

0.0500

0.1000

0.1500

0.2000

0.2500

a

-

-0.5

P[X = a]

Example

• X has a Binomial distribution with parameters n = 20 and p = 0.70

13 want We XP

13 eexact valu The XP

1643.030.070.013

20 713

Using the Normal approximation to the Binomial distribution

Where Y has a Normal distribution with:

049.230.70.20

14)70.0(20

npq

np

21

21 131213 YPXP

Hence

5.135.12 YP

049.2

145.13

049.2

14

049.2

145.12 YP

= 0.4052 - 0.2327 = 0.1725

24.073.0 ZP

Compare with 0.1643

Normal Approximation to the Binomial distribution

• X has a Binomial distribution with parameters n and p

21

21 bYaP

• Y has a Normal distribution

npq

np

correction continuity21

)()1()( bpapapbXaP

-

0.0500

0.1000

0.1500

0.2000

0.2500

a b

-

-0.5

21a 2

1b

bXaP

-

0.0500

0.1000

0.1500

0.2000

0.2500

a b

-

-0.5

21a 2

1b

21

21 bYaP

Example


1411 want We XP 1411 eexact valu The XP

614911 30.070.014

2030.070.0

11

20

)14()13()12()11( pppp

5357.01916.01643.01144.00654.0

Using the Normal approximation to the Binomial distribution

Where Y has a Normal distribution with:

049.230.70.20

14)70.0(20

npq

np

21

21 14101411 YPXP

Hence

5.145.10 YP

049.2

145.14

049.2

14

049.2

145.10 YP

= 0.5948 - 0.0436 = 0.5512

24.071.1 ZP

Compare with 0.5357

Comment:

• The accuracy of the normal appoximation to the binomial increases with increasing values of n

Example• The success rate for an Eye operation is 85%• The operation is performed n = 2000 times

Find1. The number of successful operations is

between 1650 and 1750.2. The number of successful operations is at

most 1800.

Solution


17201680 want We XP

5.17205.1679 YP

where Y has a Normal distribution with:

969.1515.85.200

1700)85.0(2000

npq

np

17201680 Hence XP

969.15

17005.1720

969.15

1700

969.15

17005.1679 YP

= 0.9004 - 0.0436 = 0.8008

28.128.1 ZP

5.17205.1679 YP

Solution – part 2.

1800 want We XP

5.1800 YP

969.15

17005.1800

969.15

1700YP

= 1.000

29.6 ZP

the normal probability distribution points of inflection 2 3 2 3 2 3 2 3

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