Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

22
Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2

Transcript of Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Page 1: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Standardizing to z

© Christine Crisp

“Teach A Level Maths”

Statistics 2

Page 2: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

We can’t have a table of probabilities for every possible mean and variance ( as we would need an infinite number of tables! )

)1,0(NSo, we always standardise to . ( Mean = 0, variance = 1 ). This is easy to do.

)110,350(~ 2NX

If X is a random variable with distribution

then, if )1,0(~ NZ110

350X

Z

The rule is“subtract the mean and divide by the standard deviation”

Since this formula holds for X, it also holds for all the values of X, given by x.

Page 3: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Using the calculator

Page 4: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

X350

x

z

110

350400 z

450 z

So, )450()400( ZPXP

Tables only give 2 d.p. for z so this is all we need.

)450( 67360 450

Z

Solution: (a)

)400( XP

x = 400, so400

)110,350(~ 2NXe.g.1 If X is a random variable with distribution

find (a) (b) )400( XP )400250( XP

Page 5: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

X

350

)110,350(~ 2NXe.g.1 If X is a random variable with distribution

find (a) (b) )400( XP )400250( XP

110

3502501z

Solution: (b) )400250( XP

910

So, )450910()400250( ZPXP

400250

110

3504002z 450

There are 2 values to convert so we use subscripts for z.

N.B. This is left of the mean so the z value will be negative.

Page 6: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

)110,350(~ 2NXe.g.1 If X is a random variable with distribution

)910()450(

818601

Solution: (b)

)450910()400250( ZPXP

450

Z

910

)910(1)910(

1814067360

18140

49220 )910()450(

find (a) (b) )400( XP )400250( XP

Page 7: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Tip: The diagrams for X and Z show the same areas so I don’t always draw both. If the question is straightforward I draw only the Z diagram but if I’m not sure what to do I’ll draw the X diagram ( and maybe the Z one as well ).

SUMMARY To use tables to solve problems, we convert

the values of the random variable X to values of the standardised normal variable using

x

z

We need to be careful not to confuse standard deviation and variance.

e.g. means s = 4.)16,20(~ NX

Page 8: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

26,29

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?Solution:Let X be the random variable “ life of battery ( hours )”

)(~ NX

Page 9: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

24

26,29

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?Solution:Let X be the random variable “ life of battery ( hours )”

)( XPWe want to find

)(~ NX

Page 10: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

24

26,29

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?Solution:Let X be the random variable “ life of battery ( hours )”

)( XPWe want to find

6

2924

830

zx 24

)830()24( ZPXPSo,)830(1 796701 20330

Approximately 20% do not last for 24 hours.

)(~ NXZ

830 0

)830(

Page 11: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Exercise

)9,15(~ NX1. If X is a random variable with distribution

2. A shop sells curtain rails labelled 90 cm. In fact the lengths are normally distributed with mean 90·2 cm. and standard deviation 0·4 cm. What percentage of the rails are shorter than 90 cm ?

find (a) (b) (c) )1611( XP )17( XP)18( XP

Page 12: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Solutions:

)9,15(~ NX(a) )18( XP

1.

13

1518

z

Z

10

)1()1( ZP)18( XP

( This is 1 standard deviation above the mean. )

84130 )18( XP

Page 13: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

)9,15(~ NX

Solutions:

1.

(b) )1611( XP

3313

15111

z 330

3

15162

z

Z

330 0331

)1611( XP )330331( ZP

)331()330(

)331(1)331( 908201

09180

0918062930)331()330( 53750

Page 14: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

)9,15(~ NX

Solutions:

1. 670

3

1517

z

Z

670 0

)670( ZP

)670(1 748601

25140

(c) )17( XP

Page 15: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Z

050

)50(1 691501

30850

2. A shop sells curtain rails labelled 90 cm. In fact the lengths are normally distributed with mean 90·2 cm. and standard deviation 0·4 cm. What percentage of the rails are shorter than 90 cm ?

Solution:

Let X be the random variable “length of rail (cm)”

)90( XPWe want to find

)40,290(~ 2 NX

5040

29090

z)50( ZP

Approximately 31% are shorter than 90 cm.

)50(

Page 16: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.
Page 17: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

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.

Page 18: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Standardizing to Z

We can’t have a table of probabilities for every possible mean and variance ( as we would need an infinite number of tables! )

)1,0(NSo, we always standardise to . ( Mean = 0, variance = 1 ). This is easy to do.

)110,350(~ 2NX

If X is a random variable with distribution

then, if )1,0(~ NZ110

350X

Z

The rule is“subtract the mean and divide by the standard deviation”

Since this formula holds for X, it also holds for all the values of X, given by x.

Page 19: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Standardizing to Z

SUMMARY

x

z

We need to be careful not to confuse standard deviation and variance.

e.g. means s = 4.)16,20(~ NX

To use tables to solve problems, we convert the values of the random variable X to values of the standardised normal variable using

Page 20: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Standardizing to Z

X350

x

z

110

350400 z

450 z

So, )450()400( ZPXP

Tables only give 2 d.p. for z so this is all we need.

)450( 67360 450

Z

Solution: (a)

)400( XP

x = 400, so400

)110,350(~ 2NXe.g.1 If X is a random variable with distribution

find (a) (b) )400( XP )400250( XP

Page 21: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Standardizing to Z

)910()450(

818601

Solution: (b)

)450910()400250( ZPXP

450

Z

910

)910(1)910(

1814067360

18140

49220 )910()450(

Page 22: Standardizing to z © Christine Crisp “Teach A Level Maths” Statistics 2.

Standardizing to Z

24

26,29

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time?Solution:Let X be the random variable “ life of battery ( hours )”

)( XPWe want to find

6

2924 z

830 )830()24( ZPXPSo,

)830(1 796701 20330 Approximately 20% do not last for 24 hours.

)(~ NXZ

830 0