standardizing to z © christine crisp teach a level maths statistics 1
TRANSCRIPT
Standardizing to Standardizing to zz
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Statistics 1Statistics 1
Standardizing to Z
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S1: Standardizing to z
AQA
Edexcel
Normal Distribution diagrams in this presentation have been drawn using FX Draw
( available from Efofex at www.efofex.com )
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.
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
Standardizing to Z
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.
Standardizing to Z
)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
Standardizing to Z
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 = 4.)16,20(~ NX
Standardizing to Z
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
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
)(~ NX
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
830
zx 24
)830()24( ZPXPSo,)830(1 796701 20330
Approximately 20% do not last for 24 hours.
)(~ NXZ
830 0
)830(
Standardizing to Z
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
Standardizing to Z
Solutions:
)9,15(~ NX
(a) )18( XP1.
13
1518
z
Z
10
)1()1( ZP)18( XP
( This is 1 standard deviation above the mean. )
84130 )18( XP
Standardizing to Z
)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
Standardizing to Z
)9,15(~ NX
Solutions:
1. 670
3
1517
z
Z
670 0
)670( ZP
)670(1 748601
25140
(c) )17( XP
Standardizing to Z
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(
Standardizing to Z
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.
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.
Standardizing to Z
SUMMARY
x
z
We need to be careful not to confuse standard deviation and variance.
e.g. means = 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
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
Standardizing to Z
)910()450(
818601
Solution: (b)
)450910()400250( ZPXP
450
Z
910
)910(1)910(
1814067360
18140
49220 )910()450(
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