Transcript 28 Standardising to Z

“Teach A Level Maths”
Statistics 1
Standardizing to z
© Christine Crisp
Standardizing to Z
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 )
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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! )
So, we always standardise to N (0, 1) .
( Mean = 0, variance = 1 ). This is easy to do.
If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
then,
Z ~ N (0, 1)
if
X  350
Z
110
Since this formula holds for X, it also holds for all the
values of X, given by x.
The rule is
“subtract the mean and divide by the standard deviation”
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (a)
z
So,
x

P ( X  400)
X
  350
x = 400, so

400  350
z
110

z  0  45
400
Tables only give 2
d.p. for z so this is
all we need.
P ( X  400)  P ( Z  0  45)
 (0  45 )
 0  6736
Z
0  45
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (b) P ( 250  X  400)
There are 2 values to
convert so we use
subscripts for z.
X
250
350 400
N.B. This is left of
the mean so the z
value will be negative.
So,
250  350
z1 
  0  91
110
400  350
z2 
 0  45
110
P(250  X  400)  P (0  91  Z  0  45)
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (b)
P(250  X  400)  P (0  91  Z  0  45)
 (0  45)  ( 0  91)
Z
( 0  91)  1  (0  91)
 0  91
 1  0  8186
 0  1814
0  45
(0  45)  ( 0  91)  0  6736  0  1814
 0  4922
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
z
x

 We need to be careful not to confuse standard
deviation and variance.
e.g.
X ~ N ( 20, 16) means  = 4.
Standardizing to Z
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 )”
 X ~ N ( 29, 6 2 )
Standardizing to Z
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 )”
 X ~ N ( 29, 6 2 )
We want to find P ( X  24)
Standardizing to Z
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 )”
 X ~ N ( 29, 6 2 )
We want to find P ( X  24)
24  29
x  24  z 
6
 0  83
So,
Z
 0  83 0
P ( X  24)  P( Z  0  83)  ( 0  83 )
 1  (0  83 )  1  0  7967  0  2033
Approximately 20% do not last for 24 hours.
Standardizing to Z
Exercise
1. If X is a random variable with distribution
X ~ N (15, 9)
find (a)
P ( X  18) (b) P (11  X  16) (c) P ( X  17)
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 ?
Standardizing to Z
Solutions:
1. X ~ N (15, 9)
(a)
P ( X  18)
18  15
z
1
3
P ( X  18)  P ( Z  1)
 (1)
Z
( This is 1 standard deviation
above the mean. )
P ( X  18)  0  8413
0
1
Standardizing to Z
Solutions:
1. X ~ N (15, 9)
(b) P (11  X
 16)
11  15
z1 
 1  33
3
Z
16  15
z2 
 0  33
3
P (11  X  16)  P (1  33  Z  0  33)
 (0  33)  ( 1  33)
( 1  33)  1  (1  33)
 1  33
 1  0  9082
 0  0918
0 0  33
(0  33)  (1  33)  0  6293  0  0918
 0  5375
Standardizing to Z
Solutions:
1. X ~ N (15, 9)
(c) P ( X  17)
 P( Z  0  67)
17  15
z
 0  67
3
Z
 1  (0  67 )
0 0  67
 1  0  7486
 0  2514
Standardizing to Z
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)”
 X ~ N (90  2, 0  4 2 )
90  90  2
z
 0  5
04
We want to find P ( X  90)
 P( Z  0  5)
 ( 0  5)
 1  (0  5)
 1  0  6915
 0  3085
Z
 05 0
Approximately 31% are shorter than 90 cm.
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! )
So, we always standardise to N (0, 1) . ( Mean = 0,
variance = 1 ). This is easy to do.
If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
then,
Z ~ N (0, 1)
if
X  350
Z
110
Since this formula holds for X, it also holds for all the
values of X, given by x.
The rule is
“subtract the mean and divide by the standard deviation”
Standardizing to Z
SUMMARY
 To use tables to solve problems, we convert the
values of the random variable X to values of the
standardised normal variable using
z
x

 We need to be careful not to confuse standard
deviation and variance.
e.g.
X ~ N ( 20, 16) means  = 4.
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (a)
z
So,
x

P ( X  400)
X
  350
x = 400, so

400  350
z
110

z  0  45
400
Tables only give 2
d.p. for z so this is
all we need.
P ( X  400)  P ( Z  0  45)
 (0  45 )
 0  6736
Z
0  45
Standardizing to Z
Solution: (b)
P(250  X  400)  P (0  91  Z  0  45)
 (0  45)  ( 0  91)
Z
( 0  91)  1  (0  91)
 0  91
 1  0  8186
 0  1814
0  45
(0  45)  ( 0  91)  0  6736  0  1814
 0  4922
Standardizing to Z
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 )”
 X ~ N ( 29, 6 2 )
We want to find P ( X  24)
24  29
z
6
 0  83
So,
Z
 0  830
P ( X  24)  P( Z  0  83)
 1  (0  83 )  1  0  7967  0  2033
Approximately 20% do not last for 24 hours.