The Normal Distribution © Christine Crisp “Teach A Level Maths” Statistics 1.

The Normal DistributionThe Normal Distribution

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Statistics 1Statistics 1

The Normal Distribution

"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

Statistics 1

AQA

Edexcel

Normal Distribution diagrams in the examples and exercises in this presentation have been drawn using FX Draw ( available from Efofex at www.efofex.com )

http://www.efofex.com/

The Normal DistributionSuppose we have a crate of apples which are to be

sorted by weight into small, medium and large. If we wanted 25% to be in the large category, we would need to know the lowest weight a “large” apple could be.

To solve a problem like this we can use a statistical model.A model often used for continuous quantities such as weight, volume, length and time is the Normal Distribution.

( Here I am using weight in the everyday sense; the quantity measured in kilograms and grams. If you are a physicist you will refer to mass. )

The Normal Distribution is an example of a probability model.


If we were to show the weights of a large number of our apples in a histogram we might get this:

There are not many very light . . . or very heavy apples.The distribution is fairly

symmetric.

Characteristics of the Normal Distribution


The Normal Distribution curve

2To fit the curve we use the mean, , and variance, , of the data. These are the parameters of the model.

),(~ 2NX

If X is the random variable “ the weight of apples”, we write

Characteristics of the Normal DistributionThe Normal distribution model is a symmetric bell-shaped curve. We fit it as closely as possible to the data.

Reminder: is the standard deviation.

The Normal DistributionCharacteristics of the Normal

Distribution

The Normal Distribution curve

For this curve we might have

)110,350(~ 2NX

The Normal distribution model is a symmetric bell-shaped curve. We fit it as closely as possible to the data.

If a question gives me the standard deviation, I often write the variance in this form instead of simplifying.


e.g.

)1,5(~ NX)1,4(~ NX

The axis of symmetry of the Normal distribution passes through the mean.


e.g.

)50,4(~ 2NX

A smaller variance “squashes” the distribution closer to the mean.

)1,4(~ NX

The Normal DistributionFinding

probabilitiesWhen we had a discrete distribution we could find a probability by using a formula.

3,2,16

)( xx

xXP for

e.g. The r.v. X has probability distribution function (p.d.f.) given by

For a continuous distribution, a probability is given by an area under the graph of the p.d.f.

1 32x

)( xXP 50

6

2P

1 2 3x

(X = )x 6

3

6

1

The Normal DistributionFor example, the probability that an apple taken

at random weighs less than 200 grams is given by the area to the left of a line through 200.

The total area gives the sum of the probabilities so equals 1.

)110,350(~ 2NX

)200( XP

Since this is a difficult function the probabilities have been worked out and listed in a table.

The p.d.f. of the Normal curve is

22

2)(

2

1)(

x

exf

The Normal DistributionFinding

Probabilities

Before we do an example, find the table of probabilities in your formulae book.

Since we may need to use these values to find others, we will always draw a sketch.

Notice the diagram at the top of the page. The shading shows that the probabilities given are always less than the z value used.

The table gives probabilities for the random variable Z where )1,0(~ NZ

( I will usually write the values to 4 d.p. )


0

)1,0(~ NZ

61

)61(

)1,0(~ NZe.g.1 If Z is a random variable with distribution

find (a) (b) (c))1( ZP )61( ZP )21( ZP

Solution: )1( ZP(a)

84130

0

)1,0(~ NZ

1

(use the table )

)61( ZP(b))61(1 ZP

is a lot to write so

we can write (1·6) which means the area to the left of 1·6.

)61( ZP



find (a) (b) (c))1( ZP )61( ZP )21( ZP

Solution: )1( ZP(a)

84130

(use the table )

)61( ZP(b)

)61(1 945201

05480 Tip: It’s useful to always check the answer: an area less than 0·5 corresponds to less than half the area and vice versa.

0

)1,0(~ NZ

61

)61(

0

)1,0(~ NZ

1


Solution:

)21( ZP(c)


find (a) (b) (c))1( ZP )61( ZP )21( ZP

0

)1,0(~ NZ

1 2 )2(

)1,0(~ NZ

0 1 2

)2(

)1(


Solution:

)21( ZP(c)

97730


find (a) (b) (c))1( ZP )61( ZP )21( ZP

0

)1,0(~ NZ

1 2 )2( )1(

84130

13600

(use the table )

)1,0(~ NZ

0 1 2

)1(

The Normal DistributionSpecial

Cases• The probability of any single value is

zero.

0)1( ZPe.g.

0

)1,0(~ NZ

1

( There is no area. )

• Also, using is the same as using <

)1()1( ZPZPe.g.

Neither property holds for discrete distributions.

The Normal DistributionSUMMARY

The Normal Distribution is continuous and symmetric about the mean.

The formula booklet gives a table of probabilities for the random variable Z where

)1,0(~ NZ

An area under the curve gives a probability.

When using the table we always draw a sketch showing the required probability.

The table gives values of . This is written as

)( zZP ).(z

z

)( zZP

0

)(z


1 10

Z

e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean.

Solution: Since we want ( the standard deviation ), the variance, , is also equal to 1. 2

1

We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1.

The easiest way to find this area is to find . . . )1(and subtract . . . )0(

1 10

Z

)11( ZP

)1,0(~ NZThe mean can be any value, so let .


1 10

Z


The easiest way to find this area is to find . . . )1(and subtract . . . )0( then multiply by

2.

1 10

Z

)11( ZP


1




1 10

Z


Can you see what equals without using the table?

)0(1 10

Z

)11( ZP

50)0(


1




1 10

Z


5000084130)0()1( 34130 341302)11( ZP 68260

1 10

Z

)11( ZP


1



The Normal Distributione.g. 2 Find the percentage of the Normal

distribution that lies within 1 standard deviation on either side of the mean.

68260)11( ZP

1 10

Z

)11( ZP

The percentage is approximately 68%.


1



The Normal DistributionExercis

e

)1,0(~ NZ

1. If Z is a random variable with distribution

2. Find the percentage of the Normal distribution that lies within (a) 2 standard deviations either side of the mean and (b) 3 standard deviations either side of the mean.

find (a) (b) (c) )451( ZP )8180( ZP)31( ZP


1(a)

)31( ZP

Solutions:

310

Z

)31( 90320

)451( ZP1(b

)

4510

Z

)451(1

926501

07350

The Normal DistributionSolutions

:

)8180( ZP1(c)

)80()81( 7881096410

17600 0

Z

8180


2 0

Z

2

2. Find the percentage of the Normal distribution that lies within (a) 2 standard deviations either side of the mean and (b) 3 standard deviations either side of the mean.

)1,0(~ NZSolution: Let

(a) We want )22( ZP

)0()2(

477250

4772502)22( ZP95450

Approximately 95% of the Normal distribution lies within 2 standard deviations of the mean. (b) The method is the same. The answer is approx.

99·8%.

50000977250


The percentages of the Normal Distribution lying within the given number of standard deviations either side of the mean are approximately:

SUMMARY

1 s.d. : 68%

2 s.d. : 95% 3 s.d. : 99·8%

68%

22

95%

33

99·8%



find (a) (b) (c))1( ZP )51( ZP )21( ZPSolution:(a) )1( ZP

Z

01

The table only gives probabilities for positive z values so we have to find an equal area that is in the table.

Z

01 1

)1()1( ZPZP)1(84130


)51( ZP

Solution:(b) Z

051

)51(1

Z

0 51

This area equals

933201 06680


)2(

841301

Solution:(c) )21( ZP Z

01 2Tip: Work out first or you could make a sign error.

)1(

)1()1(

Z

0 11

)1( )1(

15870

1587097730)1()2( So,

81860

)1(


SUMMARY To find , the area to the left of a negative

number, we use)( z

)(1)( zz N.B. The procedure for this and all other areas

involving negative values can be seen from the diagram.

NEVER try to do these questions without at least one diagram.


e)1,0(~ NZ1. If Z is a random variable with distribution

find (a) (b) (c) (d)

)31( ZP )1( ZP)212( ZP )340261( ZP

There are 2 methods of doing part (d). See if you can spot them both and use the quicker.


e)1,0(~ NZ1. If Z is a random variable with distribution


)31( ZP )1( ZP)212( ZP

Solution:(a) )31( ZP

Z

031

Z

0 31

=

)31()31( ZP

90320

)340261( ZP


e

Solution:(b) )1( ZP

Z

01

Z

0 1

=

)1(1)1( ZP 841301 15870

)1,0(~ NZ1. If Z is a random variable with distribution


)31( ZP )1( ZP)212( ZP )340261( ZP


e

Solution:

(c) )212( ZP

)212( ZP



)31( ZP )1( ZP)212( ZP

Z

02 21

0 21

)21(

02

)2

)2(1)2(

)340261( ZP

)2()21( 977301 02270

0227088490)212( ZP86220


Z

261340

Exercise



)31( ZP )1( ZP)212( ZP )340261( ZP

(d)Solution: )340261( ZP

26310 6331089620

Method 1:)340()261( Area equals

340 261


e

(d)



)31( ZP )1( ZP)212( ZP )340261( ZP

)340(1)340(

Solution: )340261( ZP

)261()340( )340261( ZP

)261(1)261( 896201

36690

10380 1038036690)340261( ZP 26310

633101

Method 2:

Z

261340


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

The Normal DistributionSUMMARY

The Normal Distribution is continuous and symmetric about the mean.

The formula booklet gives a table of probabilities for the random variable Z where

)1,0(~ NZ

An area under the curve gives a probability.

When using the table we always draw a sketch showing the required probability.

The table gives values of . This is written as

)( zZP ).(z

z

)( zZP

0

)(z

The Normal DistributionSpecial Cases• The probability of any single value is

zero.

0)1( ZPe.g.

0

Z

1

( There is no area )

• Also, using is the same as using <

)1()1( ZPZPe.g.

Neither property holds for discrete distributions.



find (a) (b) (c))1( ZP )61( ZP )21( ZP

Solution: )1( ZP(a)

84130

0

Z

1

(use the table )

)61( ZP(b)

0

Z

61

)61(1

945201 05480 0

Z

61

)61( )61( ZP


Z

0 1 2

Solution: (c)

97730

)21( ZP

0

Z

1 2

)2( )1(

84130 13600

~Z

0 1 2

)1(

)21( ZP

)2(


1 10

Z


)1,0(~ NZSolution: Let

1 10

Z

)11( ZP

We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1 where is the mean and equals zero.

5000084130)0()1( 34130

341302)11( ZP 68260 The percentage is approximately 68%.


The percentages of the Normal Distribution lying within the given number of standard deviations either side of the mean are approximately:

SUMMARY

1 s.d. : 68%

2 s.d. : 95% 3 s.d. : 99·8%

68%

22

95%

33

99·8%



find (a) (b) (c))1( ZP )51( ZP )21( ZPSolution:(a) )1( ZP

Z

01

The table only gives probabilities for positive z values so we have to find an area equal to this that is in the table.

Z

01 1

)1()1( ZPZP)1(84130


)51( ZP

Solution:(b) Z

051

)51(1

Z

0 51

This area equals

933201 06680


)2(

841301

Solution:(c) )21( ZP Z

01 2

Tip: Work out first.

)1(

)1()1(

Z

0 11

)1( )1(

15870

1587097730)1()2( So,

71860

)1(


SUMMARY To find , the area to the left of a negative

number, we use)( z

)(1)( zz N.B. The procedure for this and all other areas

involving negative values can be seen from the diagram.

NEVER try to do these questions without at least one diagram.

The Normal Distribution © Christine Crisp “Teach A Level Maths” Statistics 1.

Documents

Transcript of The Normal Distribution © Christine Crisp “Teach A Level Maths” Statistics 1.