Transcript 29a Reversing Normal calculations EDEXCEL

“Teach A Level Maths”
Statistics 1
Reversing Normal
Calculations
© Christine Crisp
Reversing Normal Calculations
Statistics 1
EDEXCEL
Normal Distribution diagrams in this presentation have been drawn using FX Draw
( available from Efofex at www.efofex.com )
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Reversing Normal Calculations
In the previous presentation there was an example where
a claim was made that batteries would last 24 hours.
However, 20% of them failed in under this time.
This manufacturer would get a lot of complaints!
However, in many industrial processes times or lengths
or weights cannot be exact and some items will fall
outside acceptable levels.
Suppose the battery manufacturer is prepared to accept
that 5% will not last as long as is claimed and so wants
to know what length of time should be claimed on the
package.
As statisticians we are being given the percentage and
want to find the corresponding value of x.
We need to reverse the process we used before.
Reversing Normal Calculations
e.g.1 A random variable X has a Normal Distribution with
mean 100 and standard deviation 15. Find the value of X
that is exceeded by 10% of the distribution.
Solution:
We want
Z
X ~ N (100 , 15 2 )
P ( X  x )  0  1. This is the same as
P( Z  z )  0  1
p  01
We now need to find z.
0
z
Instead of having to use the table backwards there is a
separate table written the other way round. It doesn’t
give as many values but contains those you are likely to
need. Find the table now. It’s called “Percentage Points
of the Normal Distribution”.
Reversing Normal Calculations
e.g.1 A random variable X has a Normal Distribution with
mean 100 and standard deviation 15. Find the value of X
that is exceeded by 10% of the distribution.
Solution:
We want
Z
X ~ N (100 , 15 2 )
P ( X  x )  0  1. This is the same as
P( Z  z )  0  1
p  01
We now need to find z.
0
z
The table shows the area to the right of z which is the
area we want, so
 z  1  2816
Reversing Normal Calculations
e.g.1 A random variable X has a Normal Distribution with
mean 100 and standard deviation 15. Find the value of X
that is exceeded by 10% of the distribution.
Solution:
We want
Z
X ~ N (100 , 15 2 )
P ( X  x )  0  1. This is the same as
P( Z  z )  0  1
p  01
z  1  2816
0
z
x
x  100
z
 1  2816 
Now we need x:

15
Rearranging:
1  2816  15  x  100
 19  224  x  100  x  119 ( 3 s. f . )
Reversing Normal Calculations
e.g.2 The running time of a batch of batteries has a normal
distribution with a mean time of 29 hours and standard
deviation 6 hours. The manufacturer can accept that 5% of
the batteries will fail to last the time he is going to claim on
the packaging. What time should this be?
Solution: Let X be the random variable “lifetime of
battery ( hours)”
2
We want
X ~ N ( 29, 6 )
P ( X  x )  0  05
Reversing Normal Calculations
e.g.2 The running time of a batch of batteries has a normal
distribution with a mean time of 29 hours and standard
deviation 6 hours. The manufacturer can accept that 5% of
the batteries will fail to last the time he is going to claim on
the packaging. What time should this be?
Solution: Let X be the random variable “lifetime of
battery ( hours)”
2
We want
Z
X ~ N ( 29, 6 )
P ( X  x )  0  05 or
p  0  95
0  05
z
0
P ( Z  z )  0  05
The largest value of p in the
tables is 0·5.
Can you see what to do?
Reversing Normal Calculations
e.g.2 The running time of a batch of batteries has a normal
distribution with a mean time of 29 hours and standard
deviation 6 hours. The manufacturer can accept that 5% of
the batteries will fail to last the time he is going to claim on
the packaging. What time should this be?
Solution: Let X be the random variable “lifetime of
battery ( hours)”
2
We want
Z
0  05
z
0
X ~ N ( 29, 6 )
P ( X  x )  0  05 or
P ( Z  z )  0  05
The largest value of p in the
tables is 0·5.
p  0  05
Can you see what to do?
ANS: We just use 0·05 BUT
the z we want is negative.
Tip: Write z  
as soon as you spot that z is
z =forget
 1  6449
negative so youSo,
don’t
the minus sign.
Reversing Normal Calculations
e.g.2 The running time of a batch of batteries has a normal
distribution with a mean time of 29 hours and standard
deviation 6 hours. The manufacturer can accept that 5% of
the batteries will fail to last the time he is going to claim on
the packaging. What time should this be?
Solution: Let X be the random variable “lifetime of
battery ( hours)”
2
We want
X ~ N ( 29, 6 )
P ( X  x )  0  05 or
Z
0  05
p  0  05
P ( Z  z )  0  05
z  1  6449
(negative)
x  29
z
  1  6449 

6
z
0
  1  6449 6  x  29
 29  9  8694  x  x  19 1
( 3 s. f . )
The time claimed should be 19 hours.
x
Reversing Normal Calculations
SUMMARY
 To solve problems where we are given a probability,
percentage or proportion we need to find values of the
random variable.
 If given a percentage or proportion we convert to a
probability and find the z value using the table called
“Percentage Points of the Normal Distribution”.
 Always draw a diagram and watch out for values of z
to the left of the mean as they will be negative.
e.g.
Z
z is negative
z
0
Write z = 
immediately.
 Use the standardizing formula to convert to x.
Reversing Normal Calculations
Exercise
1. Find the values of Z corresponding to the following:
(a) P ( Z
(c)
 z)  0  8
(b)
P ( Z  z )  0  01
P( Z  z )  0  3
(d)
P( Z  z )  0  7
Solution:
(a)
P( Z  z )  0  8
Z
p  0 2
08
0 z
z  0  8416
Reversing Normal Calculations
Solution:
(b) P ( Z
 z )  0  01
Z
p  0  01
0
(c)
z  2  3263
z
P( Z  z )  0  3
Z
p  0 3
0 3
z is negative
z  0  5244
z0
Reversing Normal Calculations
Solution:
(d) P ( Z  z )  0  7
Z
07
z is negative
Use p = 0·3,
z   0  5244
z 0
Z
p  0 3
07
0
Reversing Normal Calculations
Exercise
2.
The marks of candidates in an exam are normally
distributed with mean 50 and standard deviation 20.
(a) If 10% of candidates are to be given an A* what
mark does this correspond to?
(b) If 5% fail the exam, what is the pass mark?
Solution: Let X be the random variable “exam mark”
X ~ N (50, 20 2 )
(a) P ( X  x )  0  1  P ( Z  z )  0  1
Z
01
z  1  2816
x  50
 1  2816 
20
 x  75  6 ( 3 s. f . )
0 z
( With a mark of 75, just over 10% would get the
A* and with a mark of 76 slightly fewer. )
Reversing Normal Calculations
Exercise
2.
The marks of candidates in an exam are normally
distributed with mean 50 and standard deviation 20.
(a) If 10% of candidates are to be given an A* what
mark does this correspond to?
(b) If 5% fail the exam, what is the pass mark?
Solution: Let X be the random variable “exam mark”
(b)
X ~ N (50, 20 2 )
P( X  x )  0  05  P( Z  z )  0  05
z is negative
Z
0  05
p  0  05
z
0
z   1  6449
Reversing Normal Calculations
Exercise
2.
The marks of candidates in an exam are normally
distributed with mean 50 and standard deviation 20.
(a) If 10% of candidates are to be given an A* what
mark does this correspond to?
(b) If 5% fail the exam, what is the pass mark?
Solution: Let X be the random variable “exam mark”
(b)
X ~ N (50, 20 2 )
P( X  x )  0  05  P( Z  z )  0  05
Z
0  05
z
0
z  1  6449
x  50
p  0  05   1  6449  20
 x  17  1 ( 3 s. f . )
The pass mark would be 17.
Reversing Normal Calculations
e.g.3 The mean weight of potatoes in a batch is 150 g and
the standard deviation is 40 g. The potatoes are
graded to be put into bags. The lightest 10% are
discarded. The heaviest 20% are sorted to be sold
separately. Between what range of weights are the
potatoes in the bags?
Solution:
Let X be the random variable “weight of a potato (g)”
X ~ N (150 , 40 2 )
We have 2 percentages which we deal with separately.
P ( Z  z1 )  0  1
Z
P( Z  z 2 )  0  2
Z
0 2
01
z1
0
0 z2
Reversing Normal Calculations
P ( Z  z1 )  0  1
P( Z  z 2 )  0  2
Z
Z
01
p  01
z1
0
0 z2
z1  1  2816
z
x
p  0 2
z 2  0  8416
X ~ N (150 , 40 2 )

x1  150
 1  2816 
40
 1  2816  40  x1  150
x1  98  736
x 2  150
0  8416 
40
0  8416  40  x 2  150
x 2  183  664
The weights range from 99 g. to 184 g.
Reversing Normal Calculations
Reversing Normal Calculations
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.
Reversing Normal Calculations
e.g.1 A random variable X has a Normal Distribution with
mean 100 and standard deviation 15. Find the value of X
that is exceeded by 10% of the distribution.
Solution:
We want
Z
X ~ N (100 , 15 2 )
P ( X  x )  0  1. This is the same as
P( Z  z )  0  1
01
We now need to find z.
0
z
Instead of having to use the table backwards there is a
separate table written the other way round. It doesn’t
give as many values but contains those you are likely to
need. Find the table now. It’s called “Percentage Points
of the Normal Distribution”
Reversing Normal Calculations
The table shows the area to
the right of z which is the
area we want, so
z  1  2816
Z
p  01
0
x
z
x  100
z
 1  2816 
Now we need x:

15
Rearranging:
1  2816  15  x  100
 19  224  x  100
 x  119 ( 3 s. f . )
Reversing Normal Calculations
e.g.2 The running time of a batch of batteries has a
normal distribution with a mean time of 29 hours and
standard deviation 6 hours. The manufacturer can accept
that 5% of the batteries will not last the time written on
the packaging. What time should this be?
Solution: Let X be the random variable “lifetime of
battery ( hours)”
2
X ~ N ( 29, 6 )
We want P ( X  x )  0  05 or P ( Z  z )  0  05
The table doesn’t list values below 0·5 so we need to find
the value of z corresponding to p = 0·05 and change the sign.
Tip: Write z  
as soon
p  0  05 as you spot that z is negative so
you don’t forget the minus sign.
Z
0  05
z
0
So, z =
 1  6449
Reversing Normal Calculations
z
x

x  29
  1  6449 
6
  1  6449 6  x  29
 29  9  8694  x
 x  19 1
The time claimed should be 19 hours.
Reversing Normal Calculations
e.g.3 The mean weight of potatoes in a batch is 150 g and
the standard deviation is 40 g. The potatoes are
graded to be put into bags. The lightest 10% are
discarded. The heaviest 20% are sorted to be sold
separately. Between what range of weights are the
potatoes in the bags?
Solution:
Let X be the random variable “weight of a potato (g)”
X ~ N (150 , 40 2 )
We have 2 percentages which we deal with separately.
P ( Z  z1 )  0  1
Z
P( Z  z 2 )  0  2
Z
0 2
01
z1
0
0 z2
Reversing Normal Calculations
P ( Z  z1 )  0  1
P( Z  z 2 )  0  2
Z
Z
p  01
01
z1
0
0 z2
z1  1  2816
z
x
p  0 2
z 2  0  8416
X ~ N (150 , 40 2 )

x1  150
 1  2816 
40
 1  2816  40  x1  150
x1  98  736
x 2  150
0  8416 
40
0  8416  40  x 2  150
x 2  183  664
The weights range from 99 g. to 184 g.
Reversing Normal Calculations
SUMMARY
 To solve problems where we are given a probability,
percentage or proportion we need to find values of the
random variable.
 If given a percentage or proportion we convert to a
probability and find the z value using the table called
“Percentage Points of the Normal Distribution”.
 Always draw a diagram and watch out for values of z
to the left of the mean as they will be negative.
e.g.
Z
z is negative
z
0
Write z = 
immediately.
 Use the standardizing formula to covert to x.